GIFT  OF 


ROBBINS'S  NEW 
PLANE  GEOMETRY 


BY 

EDWARD  RUTLEDGE  ROBBINS,  A.B. 

FORMERLY   OF  LAWRENCEVILLE   SCHOOL 


AMERICAN   BOOK   COMPANY 

NEW  YORK  CINCINNATI  CHICAGO 


COPYRIGHT,  1915,  BY 
EDWARD   RUTLEDGE  ROBBINS. 


ROBBINS'S   NEW   PLANK    GEOMETRY. 
W.  P.  I 


FOR    THOSE    WHOSE    PRIVILEGE 
IT    MAY    BE    TO    ACQUIRE    A    KNOWLEDGE    OF 

GEOMETRY 

THIS   VOLUME    HAS   BEEN    WRITTEN 

AND    TO    THE    BOYS    AND    GIRLS    WHO    LEARN    THE    ANCIENT    SCIENCE 

FROM    THESE    PAGES,    AND    WHO    ESTEEM    THE    POWER 

OF    CORRECT    REASONING    THE    MORE 

BECAUSE    OF    THE    LOGIC    OF 

PURE   GEOMETRY 

THIS     VOLUME     IS     DEDICATED 


459819 


PREFACE 

THIS  New  Plane  Geometry  is  not  only  the  outgrowth,  of  the 
author's  long  experience  in  teaching  geometry,  but  has  profited 
further  by  suggestions  from -teachers  who  have  used  Robbins's 
"  Plane  Geometry  "  and  by  many  of  the  recommendations  of  the 
"National  Committee  of  Fifteen."  While  many  new  and  valu- 
able features  have  been  added  in  the  reconstruction,  yet  all  the 
characteristics  that  met  with  widespread  favor  in  the  old  book 
have  been  retained. 

Among  the  features  of  the  book  that  make  it  sound  and  teach- 
able may  be  mentioned  the  following : 

1.  The  book  has  been  written  for  the  pupil.     The  objects  sought 
in  the  study  of  Geometry  are  (1)  to  train  the  mind  to  accept 
only  those  statements  as  truth  for  which  convincing  reasons  can 
be  provided,  and  (2)  to  cultivate  a  foresight  that  will  appreci- 
ate both  the  purpose  in  making  a  statement  and  the  process  of 
reasoning  by  which  the  ultimate  truth  is  established.     Thus,  the 
study  of  this  formal  science  should  develop  in  the  pupil  the 
ability  to  pursue  argument  coherently,  and  to  establish  geometric 
truths  in  logical  order.     To  meet  the  requirements  of  the  various 
degrees  of  intellectual  capacity  and  maturity  in  every  class,  the 
reason  for  every  statement  is  not  printed  in  full  but  is  indicated 
by  a  reference.     The  pupil  who  knows  the  reason  need  not  con- 
sult the  paragraph  cited;  while  the  pupil  who  does  not  know  it 
may  learn  it  by  the  reference.     It  is  obvious  that  the  greater 
progress  an  individual  makes  in  assimilating  the  subject  and  in 
entering  into  its  spirit,  the  less  need  there  will  be  for  the  printed 
reference. 

2.  Every  effort  has  been  made  to  stimulate  the  mental  activity 
of  the  pupil.     To  compel  a  young  student,  however,  to  supply  his 


vi  PREFACE 

own  demonstrations  frequently  proves  unprofitable  as  well  as 
arduous,  and  engenders  in  the  learner  a  distaste  for  a  study  in 
which  he  might  otherwise  take  delight.  This  text  does  not  aim 
to  produce  accomplished  geometricians  at  the  completion  of  the 
first  book,  but  to  aid  the  learner  in  his  progress  throughout  the 
volume,  wherever  experience  has  shown  that  he  is  likely  to 
require  assistance.  It  is  designed,  under  good  instruction,  to 
develop  a  clear  conception  of  the  geometric  idea,  and  to  produce 
at  the  end  of  the  course  a  rational  individual  and  a  friend  of  this 
particular  science. 

3.  The  theorems  and  their  demonstrations  —  the  real  subject- 
matter  of  Geometry  —  are  introduced  as  early  in  the  study  as 
possible. 

4.  The  simple  fundamental   truths  are  explained  instead  of 
being  formally  demonstrated. 

5.  The  original  exercises  are  distinguished  by  their  abundance, 
their  practical   bearings  upon   the  affairs  of  life,  their  careful 
gradation  and  classification,  and  their  independence.     Every  ex- 
ercise can  be  solved  or  demonstrated  without  the  use  of  any  other 
exercise.     Only  the  truths  in  the  numbered  paragraphs  are  nec- 
essary in  working  originals. 

6.  The  exercises  are  introduced  as  near  as  practicable  to  the 
theorems  to  which  they  apply. 

7.  Emphasis  is  given  to  the  discussion  of  original  constructions. 

8.  The  summaries  will  be  found  a  valuable  aid  in  reviews. 

9.  The  historical  notes  give  the  pupil  a  knowledge  of  the  devel- 
opment of  the  science  of  geometry  and  add  interest  to  the  study. 

10.  The  attractive  open  page  will  appeal  alike  to  pupils  and 
to  teachers. 

The  author  sincerely  desires  to  extend  his  thanks  to  those 
friends  and  fellow  teachers  who,  by  suggestion  and  encourage- 
ment, have  inspired  him  in  the  preparation  of  these  pages. 

EDWARD   K.    BOBBINS. 


CONTENTS 

INTRODUCTION 

PAGE 

DEFINITIONS 1 

ANGLES 2 

TRIANGLES 4 

CONGRUENCE      ......        5 

SYMBOLS      ............  6 

AXIOMS 6 

POSTULATES .  7 

EXERCISES 9 

BOOK  I.    ANGLES,  LINES,  RECTILINEAR  FIGURES 

PRELIMINARY  THEOREMS  .        .        . 13 

THEOREMS  AND  DEMONSTRATIONS 15 

TRIANGLES 15 

PARALLEL  LINES •  .        .20 

QUADRILATERALS 47 

POLYGONS 60 

SYMMETRY 65 

CONCERNING  ORIGINAL  EXERCISES  .        .        .        ....  68 

SUMMARY.     GENERAL  DIRECTIONS  FOR  ATTACKING  ORIGINALS   .  68 

ORIGINAL  EXERCISES 70 

BOOK  II.    THE  CIRCLE 

DEFINITIONS 75 

PRELIMINARY  THEOREMS 77 

THEOREMS  AND  DEMONSTRATIONS 78 

SUMMARY 94 

ORIGINAL  EXERCISES        .                95 

vii 


vi  PREFACE 

own  demonstrations  frequently  proves  unprofitable  as  well  as 
arduous,  and  engenders  in  the  learner  a  distaste  for  a  study  in 
which  he  might  otherwise  take  delight.  This  text  does  not  aim 
to  produce  accomplished  geometricians  at  the  completion  of  the 
first  book,  but  to  aid  the  learner  in  his  progress  throughout  the 
volume,  wherever  experience  has  shown  that  he  is  likely  to 
require  assistance.  It  is  designed,  under  good  instruction,  to 
develop  a  clear  conception  of  the  geometric  idea,  and  to  produce 
at  the  end  of  the  course  a  rational  individual  and  a  friend  of  this 
particular  science. 

3.  The  theorems  and  their  demonstrations — the  real  subject- 
matter  of  Geometry  —  are  introduced  as  early  in  the  study  as 
possible. 

4.  The  simple  fundamental   truths  are  explained  instead  of 
being  formally  demonstrated. 

5.  The  original  exercises  are  distinguished  by  their  abundance, 
their  practical   bearings  upon   the  affairs  of   life,  their  careful 
gradation  and  classification,  and  their  independence.     Every  ex- 
ercise can  be  solved  or  demonstrated  without  the  use  of  any  other 
exercise.     Only  the  truths  in  the  numbered  paragraphs  are  nec- 
essary in  working  originals. 

6.  The  exercises  are  introduced  as  near  as  practicable  to  the 
theorems  to  which  they  apply. 

7.  Emphasis  is  given  to  the  discussion  of  original  constructions. 

8.  The  summaries  will  be  found  a  valuable  aid  in  reviews. 

9.  The  historical  notes  give  the  pupil  a  knowledge  of  the  devel- 
opment of  the  science  of  geometry  and  add  interest  to  the  study. 

10.  The  attractive  open  page  will  appeal  alike  to  pupils  and 
to  teachers. 

The  author  sincerely  desires  to  extend  his  thanks  to  those 
friends  and  fellow  teachers  who,  by  suggestion  and  encourage- 
ment, have  inspired  him  in  the  preparation  of  these  pages. 

EDWARD  R.    ROBBINS. 


CONTENTS 

INTRODUCTION 

PAGE 

DEFINITIONS 1 

ANGLES 2 

TRIANGLES 4 

CONGRUENCE      ......        5 

SYMBOLS      . 6 

AXIOMS        . 6 

POSTULATES 7 

EXERCISES 9 

BOOK  I.    ANGLES,  LINES,  RECTILINEAR  FIGURES 

PRELIMINARY  THEOREMS 13 

THEOREMS  AND  DEMONSTRATIONS 15 

TRIANGLES 15 

PARALLEL  LINES •  .        .20 

QUADRILATERALS       ..........  47 

POLYGONS 60 

SYMMETRY 65 

CONCERNING  ORIGINAL  EXERCISES  .        .        .        ....  68 

SUMMARY.     GENERAL  DIRECTIONS  FOR  ATTACKING  ORIGINALS   .  68 

ORIGINAL  EXERCISES 70 

BOOK  II.    THE  CIRCLE 

DEFINITIONS 75 

PRELIMINARY  THEOREMS 77 

THEOREMS  AND  DEMONSTRATIONS 78 

SUMMARY 94 

ORIGINAL  EXERCISES        .                95 

vii 


viii  CONTENTS 

PAGE 

KINDS  OF  QUANTITIES.     MEASUREMENT 96 

ORIGINAL  EXERCISES 109 

Loci 114 

ORIGINAL  EXERCISES  ON  Loci 115 

CONSTRUCTION  PROBLEMS 117 

ANALYSIS 131 

ORIGINAL  CONSTRUCTION  PROBLEMS 132 

BOOK  III.    PROPORTION.    SIMILAR  FIGURES 

DEFINITIONS 143 

THEOREMS  AND  DEMONSTRATIONS   . 144 

CONCERNING  ORIGINALS    . 175 

ORIGINAL  EXERCISES  (NUMERICAL) 176 

SUMMARY 180 

ORIGINAL  EXERCISES  (THEOREMS) 180 

CONSTRUCTION  PROBLEMS 186 

ORIGINAL  CONSTRUCTION  PROBLEMS       .        .        .        .    •     .        .  190 

BOOK  IV.    AREAS 

THEOREMS  AND  DEMONSTRATIONS 193 

FORMULAS 207 

ORIGINAL  EXERCISES  (NUMERICAL) 210 

CONSTRUCTION  PROBLEMS 213 

ORIGINAL  CONSTRUCTION  PROBLEMS 221 

BOOK  V.     REGULAR  POLYGONS.    CIRCLES 

THEOREMS  AND  DEMONSTRATIONS 225 

ORIGINAL  EXERCISES  (THEOREMS) 239 

CONSTRUCTION  PROBLEMS 242 

FORMULAS 245 

ORIGINAL  EXERCISES  (NUMERICAL) 248 

ORIGINAL  CONSTRUCTION  PROBLEMS 253 

MAXIMA  AND  MINIMA 254 

ORIGINAL  EXERCISES 259 

INDEX  .        .        .261 


PLANE  GEOMETRY 


INTRODUCTION 

1.  Geometry  is  a  science  which  treats  of  the  measure- 
ment of  magnitudes. 

2.  A  point  is  that  which  has  position  but  not  magnitude. 

3.  A  line  is  that  which  has  length  but  no  other  magni- 
tude. 

4.  A  straight  line  is  a  line  which  is  determined  (fixed  in 
position)  by  any  two  of  its  points.     That  is,  two  lines  that 
coincide   entirely,  if  they  coincide  at  any  two  points,  are 
straight  lines. 

5.  A  rectilinear  figure  is  a  figure  containing  straight  lines 
and  no  others. 

6.  A  surface  is  that  which  has  length  and  breadth  but  no 
other  magnitude. 

7.  A  plane  is  a  surface  in  which  if  any  two  points  are 
taken,  the  straight  line  connecting  them  lies  wholly  in  that 
surface. 

8.  Plane  Geometry  is  a  science  which  treats  of  the  proper- 
ties of  magnitudes  in  a  plane. 

9.  A  solid  is  that  which  has  length,  breadth,  and  thick- 
ness.    A  solid  is  that  which  occupies  space. 

10.   Boundaries.    The  boundaries  (or  boundary)  of  a  solid 
are  surfaces.     The  boundaries  (or  boundary)  of  a  surface 

1 


PLANE  GEOMETRY 

;aJr,e'  'filler :  -'T-he  boundaries  of  a  line  are  points.  These 
boundaries  can  be  no  part  of  the  things  they  limit.  A  sur- 
face is  no  part  of  a  solid ;  a  line  is  no  part  of  a  surface ;  a 
point  is  no  part  of  a  line. 

11.  Motion.  If  a  point  moves,  its  path  is  a  line.  Hence, 
if  a  point  moves,  it  generates  (describes  or  traces)  a  line ;  if 
a  line  moves  (except  upon  itself),  it  generates  a  surface ; 
if  a  surface  moves  (except  upon  itself),  it  generates  a  solid. 

NOTE.    Unless  otherwise  specified  the  word  "  line  "  means  straight  line. 

ANGLES 


ANGLE          ADJACENT     VERTICAL  ANGLES     RIGHT  ANGLES 
ANGLES  PERPENDICULAR 

12.  A  plane  angle  is  the  amount  of  divergence  of  two 
straight   lines   that   meet.     The   lines   are  called  the  sides 
of  the  angle.     The  vertex  of  an  angle  is  the  point  at  which 
the  lines  meet. 

13.  Adjacent  angles  are  two  angles  that  have  the  same 
vertex  and  a  common  side  between  them. 

14.  Vertical  angles  are  two  angles-  that  have  the  same 
vertex,  the  sides  of  one  being  prolongations  of  the  sides  of 
the  other. 

16.  If  one  straight  line  meets  another  and  makes  the  ad- 
jacent angles  equal,  the  angles  are  right  angles. 

16.  One  line  is  perpendicular  to  another  if  they  meet  at 
right  angles.  Either  line  is  perpendicular  to  the  other.  The 
point  at  which  the  lines  meet  is  the  foot  of  the  perpendicular. 
Oblique  lines  are  lines  that  meet  but  are  not  perpendicular. 


INTRODUCTION  .  3 

17.  A  straight  angle  is  an  angle  whose  sides  lie  in  the 
same  straight  line,  but  extend  in  opposite  directions  from 
the  vertex. 


OBTUSE  ACUTE          COMPLEMENTARY  SUPPLEMENTARY 

ANGLE  ANGLE  ANGLES  ANGLES 

18.  An  obtuse  angle  is  an  angle  that  is  greater  than  a 
right  angle.     An  acute  angle  is  an  angle  that  is  less  than 
a  right  angle.     An  oblique  angle  is  any  angle  that  is  not  a 
right  angle. 

19.  Two  angles  are  complementary  if  their  sum  is  equal  to 
one  right  angle.     Two  angles  are  supplementary  if  their  sum 
is  equal  to  two  right  angles.     Thus,  the  complement  of  an 
angle  is  the  difference  between  one  right  angle  and  the  given 
angle.     The  supplement  of  an  angle  is  the  difference  between 
.two  right  angles  and  the  given  angle. 

20.  A   degree   is   one   ninetieth  of   a  right  angle.     The 
degree  is  the  familiar  unit  used  in  measuring  angles.     It  is 
evident   that   there  are  90°  in  a  right  angle ;  180°  in  two 
right  angles,  or  a  straight  angle  ;  360°  in  four  right  angles. 

There  are  60  minutes  (60')  in  one  degree,  and  60  seconds  (60")  in  one 
minute. 

21.  Parallel  lines  are  straight  lines  that  lie  in  the  same 
plane  and  that  never  meet,  however  far  they  are  extended  in 
either  direction. 

22.  Notation.     A  point  is  usually  denoted  by  a  capital  letter,  placed 
near  it.     A  line  is  denoted  by  two  capital  letters,  placed  one  at  each  end, 
or  one  at  each  of  two  of  its  points.     Its  length  is  sometimes  represented 
advantageously  by  a  small  letter  written  near  it.     Thus,  the  line  AB ; 
the  line  RS\  the  line  m. 

R  S  m 


4  PLANE   GEOMETRY 

There  are  various  ways  of  naming  angles.  Sometimes  three  capital 
letters  are  used,  one  on  each  side  of  the  angle  and  one  at  the  vertex ; 
sometimes  a  small  letter  or  a  figure  is  placed  within  the  angle.  The 
symbol  for  angle  is  Z. 


M 

ZAMXo* 
/.XMA  on  AM 


X  O  C 

Z.a  AND  Z  .BOO, 
NOT  ZO 


In  naming  an  angle  by  the  use  of  three  letters,  the  vertex  letter  is  al- 
ways placed  between  the  others.  Thus  the  A  above  are  Z.AMX  or 
Z  KM  A,  Z  a,  Z  BOC,  Z.x,/.APR,Z.  APS,  Z  BPR,  Z  TPB,  Z  5,  etc. 

In  the  above  figure  Z  x  =  Z  5.  The  size  of  an  angle  depends  on  the 
amount  of  divergence  between  its  sides,  and  not  upon  their  length. 

An  angle  is  said  to  be  included  by  its  sides.  An  angle  is 
bisected  by  a  line  drawn  through  the  vertex  and  dividing 
the  angle  into  two  equal  angles. 


TRIANGLES 

23.  A  triangle  is  a  portion  of  a  plane  bounded  by  three 
straight  lines.  These  lines  are  the  sides.  The  vertices  of  a 
triangle  are  the  three  points  at  which  the  sides  intersect. 
The  angles  of  a  triangle  are  the  three  angles  at  the  three 
vertices.  Each  side  of  a  triangle  has  two  angles  adjoining 
it.  The  symbol  for  triangle  is  A. 


ISOSCELES  A         EQUILATERAL  A         BIGHT  A         OBTUSE  A  ACUTE  A 

EQUIANGULAR  A  SCALENE  & 


INTRODUCTION  5 

The  base  of  a  triangle  is  the  side  on  which  the  figure  ap- 
pears to  stand.  The  vertex  of  a  triangle  is  the  vertex  op- 
posite the  base.  The  vertex  angle  is  the  angle  opposite  the 
base. 

24.  Kinds  of  triangles  : 

A  scalene  triangle  is  a  triangle  no  two  sides  of  which  are  equal. 
An  isosceles  triangle  is  a  triangle  two  sides  of  which  are  equal. 
An  equilateral  triangle  is  a  triangle  all  sides  of  which  are  equal. 
A  right  triangle  is  a  triangle  one  angle  of  which  is  a  right  angle. 
An  obtuse  triangle  is  a  triangle  one  angle  of  which  is  an  obtuse  angle, 
An  acute  triangle  is  a  triangle  all  angles  of  which  are  acute  angles. 
An  equiangular  triangle  is  a  triangle  all  angles  of  which  are  equal. 

25.  The  hypotenuse  of  a  right  triangle  is  the  side  opposite 
the  right  angle.    The  sides  forming  the  right  angle  are  called 
legs. 

CONGRUENCE 

26.  Two  geometric  figures  are  said  to  be   equal  if   they 
have  the  same  size  or  magnitude. 

Two  geometric  figures  are  said  to  be  congruent  if,  when 
one  is  superposed  upon  the  other,  they  coincide  in  all  respects. 

The  corresponding  parts  of  congruent  figures  are  equal, 
and  are  called  homologous  parts. 

27.  Homologous  parts  of  congruent  figures  are  equal. 

If  the  triangles  DEF  and  HIJ  are  congruent, 

Z.D\$>  homologous  to  and  =  to  Z.  H ; 

DE  is  homologous  to  and  =  to  HI; 

Z  E  is  homologous  to  and  =  to  ^  /; 

EF  is  homologous  to  and  =  to  IJ. 

NOTE.  Congruent  figures  have  the 
same  shape  as  well  as  the  same  size, 
whereas  equal  figures  do  not  necessarily  have  the  same  shape. 


Ex.1.  What  is  the  complement  of  an  angle  of  35°?  48°?  80°? 
75°  50'  ?  8°  20'  ? 

Ex.  2.  What  is  the  supplement  of  an  angle  of  100°?  50°?  148°? 
121°  30'?  10°  40'? 


6 


PLANE   GEOMETRY 


28.  A  curve  or  curved  line,  is  a  line  no  part  of  which  is 
straight. 

A  circle  is  a  plane  curve  all  points  of  which  are  equally 
distant  from  a  point  in  the  plane,  called  the  center. 

An  arc  is  any  part  of  a  circle. 

A  radius  is  a  straight  line  from  the  center  to  any  point  of 
the  circle. 

A  diameter  is  a  straight  line  containing  the  center  and 
having  its  extremities  in  the  circle. 

The  length  of  the  circle  is  called  the  circumference. 

29.  Symbols.     The  usual  symbols  and  abbreviations  em- 
ployed in  geometry  are  the  following  : 

+  plus. 

—  minus. 

=  equals,  is  equal  to, 
equal. 

=£  does  not  equal. 

^  congruent,  or  is  con- 
gruent to. 

>   is  greater  than. 

<   is  less  than. 

.'.  hence,  therefore, 
consequently. 

JL    perpendicular. 

Js    perpendiculars. 

AXIOM,   POSTULATE,   AND  THEOREM 

30.  An  axiom  is  a  statement  admitted  without  proof  to  be 
true.     It  is  a  truth,  received  and  assented  to  immediately. 

31.  AXIOMS. 

1.  Magnitudes  that  are  equal  to  the  same  thing,  or  to  equals, 
are  equal  to  each  other. 

2.  If  equals  are  added  to,  or  subtracted  from,  equals,  the  results 
are  equal. 

3.  If  equals  are  multiplied  by,  or  divided  by,  equals,  the  results 
are  equal. 

[Doubles  of  equals  are  equal;  halves  of  equals  are  equal.] 


O  circle. 

Ax. 

axiom. 

©  circles. 

Hyp. 

hypothesis. 

Z    angle. 

comp. 

complementary. 

A    angles. 

supp. 

supplementary. 

rt.  Z  right  angle. 

Const. 

construction. 

rt.  A  right  angles. 

Cor. 

corollary. 

A  triangle. 

St. 

straight. 

&   triangles. 

rt. 

right. 

rt.  &  right  triangles. 

Def. 

definition. 

II     parallel. 

alt.    . 

alternate. 

Us    parallels. 

int. 

interior. 

d  parallelogram. 

ext. 

exterior. 

£17  parallelograms. 

INTRODUCTION  7 

4.  The  whole  is  equal  to  the  sum  of  all  of  its  parts. 

5.  The  whole  is  greater  than  any  of  its  parts. 

6.  A  magnitude  may  be  displaced  by  its  equal  in  any  process. 

[Briefly  called  "substitution."] 

7.  If  equals  are  added  to,  or  subtracted  from,  unequals,  the 
results  are  unequal  in  the  same  order. 

8.  If  unequals  are  added  to  unequals  in  the  same  order,  the 
results  are  unequal  in  that  order. 

9.  If  unequals  are  subtracted  from  equals,  the  results  are  un- 
equal in  the  opposite  order. 

10.  Doubles  or  halves  of  unequals  are  unequal  in  the  same  order. 
Also,  unequals   multiplied   by  equals   are  unequal  in  the  same 
order. 

11.  If  the  first  of  three  magnitudes  is  greater  than  the  second, 
and  the  second  is  greater  than  the  third,  the  first  is  greater  than 
the  third. 

12.  A  straight  line  is  the  shortest  line  that  can  be  drawn  be- 
tween two  points. 

13.  Only  one  line  can  be  drawn  through  a  point  parallel  to  a 
given  line. 

14.  A  geometrical  figure  may  be  moved  from  one  position  to 
another  without  any  change  in  form  or  magnitude. 

32.  A  postulate  is  something  required  to  be  done,  the  pos- 
sibility of  which  is  admitted  without  proof. 

33.  POSTULATES. 

1.  It  is  possible  to  draw  a  straight  line  from  any  point  to  any 
other  point. 

2.  It  is  possible  to  extend  (prolong  or  produce)  a  straight  line 
indefinitely,  or  to  terminate  it  at  any  point. 


8  PLANE   GEOMETRY 

34.  A  geometric  proof  or  demonstration  is  a  logical  course 
of  reasoning  by  which  a  truth  becomes  evident. 

35.  A  theorem  is  a  statement  that  requires  proof. 

In  the  case  of  the  preliminary  theorems  which  follow,  the 
proof  is  very  simple ;  but  as  these  theorems  are  not  admitted 
without  proof  they  cannot  be  classified  with  the  axioms. 

A  corollary  is  a  truth  immediately  evident,  or  readily  es- 
tablished from  some  other  truth  or  truths. 

A  proposition,  in  geometry,  is  the  statement  of  a  theorem 
to  be  proved  or  a  problem  to  be  solved. 


Ex.  1.   Draw  an  Z  ABC.     In  /.  ABC  draw  line  BD.    . 
What  does  Z  ABD  +  Z.  DEC  equal  ? 
What  does  Z  ABC  -  ZABD  equal? 

Ex.  2.   In  a  rt.  Z.ABC  draw  line  BD. 
If  ZABD  =  25°,  how  many  degrees  are  there  in  Z  DB C  ? 
How  many  degrees  are  there  in  the  complement  of  an  angle  of  38°  ? 
How  many  degrees  are  there  in  the  supplement  ? 

Ex.  3.   Draw  a  straight  line  AB  and  take  a  point  X  on  it. 
What  line  does  AX  +  BX  equal? 
What  line  does  AB  -  BX  equal? 

Ex.  4.  Draw  a  straight  line  AB  and  prolong  it  to  X  so  ih&iBX  =  AB. 
Prolong  it  so  that  AB  =  AX. 

Historical  Note.  Probably  as  early  as  3000  B.C.  the  Egyptians  had 
some  knowledge  of  geometric  truths.  The  construction  of  the  great 
pyramids  required  an  acquaintance  with  the  relations  of  geometry. 
This  knowledge,  however  vague  it  may  have  been,  was,  according  to 
Herodotus,  employed  in  determining  the  amount  of  land  washed  away 
by  the  river  Nile,  during  the  reign  of  Rameses  II  (1400  B.C.). 

The  Greeks,  however,  were  the  first  to  study  geometry  as  a  logical 
science.  They  enunciated  theorems  and  demonstrated  them,  they  pro- 
pounded problems  and  solved  them  as  early  as  300  B.C.,  and,  in  a  crude 
way,  two  or  three  centuries  earlier.  To  them  belongs  the  credit  of  estab- 
lishing a  logical  system  of  geometry  that  has  survived,  practically  un- 
changed, for  twenty  centuries. 


INTRODUCTION 


B 


EXERCISES  EMPLOYING  THE  TWO   INSTRUMENTS   OF 
GEOMETRY 

Aside  from  pencil  and  paper,  the  only  instruments 
necessary  for  the  construction  of  geometrical  dia- 
grams are  the  ruler  and  the  compasses. 

Ex.  1.  It  is  required  to  draw  an  equilateral  tri- 
angle upon  a  given  line  as  base. 

Suppose  AB  is  the  given  base. 

Required  to  draw  an  equilateral  A  upon  it. 

Using  A  as  a  center  and  AB  as  a  radius,  draw  an  arc.     Using  B 
as  a  center  and  AB  as  a  radius,  draw  another  arc  cutting  the  first  one 
at  C.     Draw  AC  and  BC.     The  &ABC  is  an  equi- 
lateral A,  and  AB  is  its  base. 

Ex.  2.  It  is  required  to  draw  a  triangle  having  its 
three  sides  each  equal  to  a  given  line. 

Suppose  the  three  given  lines  are  a,b,c. 

Required  to  draw  a  A  having  for  its  sides  lines 
equal  to  a,  &,  c-,  respectively. 

Draw  a  line  RS  —  to  a.  Using  R  as  a  center  and 
b  as  a  radius,  draw  an  arc.  Using  S  as  a  center 
and  c  as  a  radius,  draw  another  arc  cutting  the  first 
arc  at  T.  Draw  straight  lines  R  T  and  ST.  A  RST 
is  the  A  whose  three  sides  are  equal  to  the  lines 
a,  b,  c,  respectively. 

Ex.  3.  It  is  required  to  find  the  midpoint  of  a 
given  straight  line. 

Given  the  straight  line  AB. 

Required  to  find  its  midpoint. 

Using  A  and  B  as  centers  and  a  radius  sufficiently              /  j    \ 
long,  draw  two  arcs,  intersecting  at  P  and  Q.  A—I — r^-j B 

Draw  the  straight  line  PQ  cutting  AB   at  M.  \    I    / 

Point  M  is  the  midpoint  of  A  B.  \  \  / 

Ex.  4.   It  is  required  to  draw  a  perpendicular  to  a  '"Qx 

line  from  a  point  within  the  line. 

Given  the  line  CD  and  point  P  in  it.  ..K 

Required  to  construct  a  _L  to  CD,  at  P. 

Using  P  as  center  and  any  radius,  draw  two  arcs    ^         /     I     \     p 
cutting  CD  at  E  and  F.     Now  using  E  and  F  as 
centers  and  a  radius  greater  than  before,  draw  two  arcs  intersecting  at  K. 
Draw  KP.     This  line  KP  is  _L  to  CD  at  P. 

BOBBINS'S  NEW  PLANE  GEOM. — 2 


10 


PLANE   GEOMETRY 


P 

L 


Ex.  6.    It  is  required  to  draw  a  perpendicular  to  a 
line  from  a  point  without  the  line. 

Given  line  AB  and  point  P,  without  it. 

Required  to  draw  a  _L  to  A  B  from  P. 

Using  P  as  center  and  a  sufficient  radius,  draw    A- 
an  arc  cutting  AB  at  C  and  D.     Now  using  C  and 
D  as  centers  and  a  sufficient  radius,  draw  two   arcs  intersecting  at  E. 
Draw  PE,  meeting  AB  at  R.     PR  is  the  required  _L  to  AB  from  P. 

Ex.  6.   It  is  required  to  bisect  a  given  angle. 

Given  the  Z  ABC. 

Required  to  bisect  it. 

Using  vertex  B  as  a  center  and  any  radius,  draw 
arc  DE  cutting  BC  at  D  and  BA  at  E. 

Using  D  and  E  as  centers-and  a  sufficient  radius,  draw  arcs  intersect- 
ing at  F.  Draw  straight  line  BF.  BF  bisects  the  Z  ABC. 

Ex.  7.  It  is  required  to  cpnstruct,  at  a  given  point  on  a  given  line,  an 
angle  equal  to  a  given  angle. 

Given   line   DE,    point  D   in  it, 
and  Z  B. 

Required  to  construct  an  Z  at  D, 
equal  to  Z  5. 

Using  .B  as  a  center  and  with  any  two  distances  as  radii,  draw  an  arc 
cutting  AB  at  F  and  another  cutting  BC  at  G. 

Using  D  as  a  center  and  the  same  radii  as  before,  draw  one  arc,  and 
another  arc  cutting  DE  at  /. 

Draw  the  straight  line  FG.  Using  /  as  a  center  and  FG  as  a  radius, 
draw  an  arc  cutting  a  former  arc  at  H.  Draw  the  straight  lines  HJ  and 
DHK. 

Now  the  Z  KDE  =  Z  B. 

Ex.  8.   By  the  use  of  ruler  and  compasses,  draw  the  following  figures : 


G 


Ex.  9.    Does  it  make  any  difference  in  these  exercises,  which  lines  are 
drawn  first?     In  Ex.  7  and  Ex.  8  explain  the  order  of  the  lines  drawn. 


INTRODUCTION  11 

Ex.  10.    Using  the  compasses  only,  draw  the  following  figures: 


Ex.  11.    Draw  the  following  figures : 


Ex.  12.    Draw  the  first  of  each  of  these  three  pairs  of  figures. 

Can  you  explain  the  construction  of  the  second  figure  in  each  pair  ? 


12 


PLANE   GEOMETRY 


In  this  figure,  ABCD  is  a  square.  On 
the  sides  are  measured  the  equal  dis- 
tances AE  and  BF,  and  CG  and  DH ;  ( 
then  the  lines  AG,  BH,  CE,  and  DF,  are 
drawn  intersecting  at  a,  b,  c,  d.  The 
figure  abed  is  also  .a  square.  This  figure 
is  the  basis  of  an  Arabic  design  used  for 
parquet  floors,  etc. 

In  this  figure,  which  is  the  basis  of  a 
mosaic  floor  design,  the  radii  of  all  com- 
plete circles  equal  one  fourth  of  the  side 
of  the  square  ABCD.  The  radii  of  the 
semicircles  GflJ,  IKR,  etc.,  equal  one 
eighth  of  the  side  of  the  square. 

In  this  figure  ABD  is  an  equilateral  arch? 
and  CD  is  its  altitude.  The  several  cen- 
ters used  are  A,  of  arc  BD  and  arc  CE ;  5, 
of  arc  AD  and  CF ;  (7,  of  arcs  AE  and  BF. 

This  figure  is  the  basis  of  a  common 
Gothic  window  design. 

NOTE.  The  letters  "  Q.E.D."  are  often  annexed  at  the  end  of  a  demon- 
stration and  stand  for  "quod  erat  demonstrandum,"  which  means,  "  which 
was  to  be  proved." 


BOOK  I 

ANGLES,  LINES,  RECTILINEAR  FIGURES 

PRELIMINARY  THEOREMS 

36.  A  right  angle  is  equal  to  half  a  straight  angle. 

Because  of  the  definition  of  a  right  angle. 

37.  A  straight  angle  is  equal  to  two  right  angles.  (36.) 

38.  Two  straight  lines  can  intersect  in  only  one  point. 

Because   they  would   coincide  entirely  if  they  had  two 
common  points.  (4.) 

• 

39.  Only  one  straight  line  can  be  drawn  between  two  points. 

(40 

40.  A  definite  (limited  or  finite)  straight  line  can  have  only 
one  midpoint. 

Because  the  halves  of  a  line  are  equal. 

41.  All  straight  angles  are  equal. 

Because  they  can  be  made  to  coincide.  (26.) 


42.  \  All  right  angles  are  equal. 

ThejKare  halves  of  straight  angles.  (36.) 

.•.  they  are  equal.  (Ax.  3.) 

43.  Only  one  perpendicular  to  a  line  can  be  drawn  from  a 

point  in  the  line. 

These  right  angles  would  not  be  equal  if  there  were  two 
perpendiculars.  (42.) 

13 


14  BOOK   I.  "PLANE   GEOMETRY 

44.  If  two  adjacent  angles  have  their  exterior  sides  in  a 
straight  line,  they  are  supplementary. 

Because  they  together  equal 
two  rt.  A.  (19.) 

45.  If  two  adjacent  angles 
are  supplementary,  their  exte- 
rior   sides    are    in    the    same      _ 
straight  line. 

Because  their  sum  is  two  rt.^  (19)  ;  or  a  straight  Z  (37) . 
Hence  the  exterior  sides  are  in  the  same  straight  line  (17). 

46.  The  sum  of  all  the  angles  on  one  side  of  a  straight  line 
at  a  point  equals  two  right  angles. 

(Ax.  4  and  37.) 

47.  The  sum  of  all  the  angles 
about  a  point  in  a  plane  is  equal 

to  four  right  angles.  (46.)    

48.  Angles  that  have  the  same  complement  are  equal.     Or, 
complements  of  the  same  angle,  or  of  equal  angles,  are  equal. 

Because  equal  angles  subtracted  from  equal  right  angles 
leave  equal  angles.  (Ax.  2.) 

49.  Angles  that  have  the  same  supplement  are  equal.     Or, 
supplements  of  the  same  angle,  or  of  equal  angles,  are  equal. 

(Ax.  2.) 

50.  If  two  angles  are  equal  and  supplementary,  they  are 
right  angles. 

Each  is  half  a  straight  Z  ;  .'.  each  is  a  rt. 


XOTE.     A  reference  number  usually  indicates  only  the  statement  in 
full  face  type  in  the  section  referred  to.     In  giving 
demonstrations  the  pupil  should  quote  the  correct 
reason  for  each  statement. 


Ex.   The  bisector^  of  two  supplementary  adja- 
cent angles  are  perpendicular  to  each  other. 


TRIANGLES  15 

THEOREMS  AND   DEMONSTRATIONS 
PROPOSITION  I.     THEOREM 

51.   If  two  straight  lines  intersect,  the 
vertical  angles  are  equal.  A 


Given :  Lines  AB  and  LM  intersecting 

at  o,  A  AOM  and  BOL,  a  pair  of  vertical  A. 

To  Prove :  Z  AOM  =  Z  BOL. 

Proof :          A  AOM  and  MOB  are  supplementary  (44). 

A  MOB  and  BOL  are  supplementary  (44). 

.'./-AOM  =  Z  BOL.  (49.) 

Q.E.D. 

PROPOSITION  II.     THEOREM 

52.  Two  triangles  are  congruent  if  two  sides  and  the  in- 
cluded angle  of  one  are  equal  respectively  to  two  sides  and  the 
included  angle  of  the  other. 


A  B   R  S 

Given :  A  ABC  and  RST ;  AB  =  RS;  AC=KT\  Z  A  =  Z  R. 
To  Prove :  A  ABC  is  congruent  to  A  RST. 
Proof :  Place  the  A  ABC  upon  the  A  RST  so  that  Z  A  coin- 
cides with  its  equal  Z  R. 

AB  falls  upon  RS  and  point  B  upon  S  (AB  =  RS~). 

AC  falls  upon  RT  and  point  C  upon  T  (AC  =  RT). 

.*.  BC  coincides  with  8T  (39). 

/.the  A  coincide  and  are  congruent  (26). 

Q.E.D. 


16 


BOOK   I.     PLANE   GEOMETRY 


53.   COROLLARY.    Two  right  triangles  are  congruent  if  two 
legs  of  one  are  equal  respectively  to  two  legs  of  the  other. 

This  is  a  corollary  following  immediately  from  52. 


B 


Ex.  1.   If  two  triangles  have  two  sides  of  one  equal  to  two  sides  of  the 
other,  are  the  triangles  necessarily  congruent? 

Ex.  2.  Illustrate  your  answer  to  Ex.  1  by  draw- 
ing two  triangles. 

Ex.  3.  Find  the  distance  AB  if  there  is  an  ob- 
struction between  A  and  B. 

Method.  Take  a  convenient  point  P,  from 
which  A  and  B  are  accessible.  Measure  AP  and  in  same  straight  line 
mark  point  R  such  that  PR  =  A  P.  Similarly  find  point  S.  Show  that 
the  two  &  ABP  and  SRP  are  congruent,  therefore 
the  length  of  AB  may  be  found  by  measuring  RS. 

Ex.  4.  Prove  that  a  point,  P,  in  the  perpendicular 
bisector  MC  of  a  line  AB  is  equally  distant  from 
the  ends  of  the  line  AB. 

(Show  that  the  &  AMP  and  BMP  are  congruent, 
(1)  by  using  52;  and  (2)  by  using  53.) 

Ex.6.  If  the  line  AC  bisects  ZBAD,  and 
BA  =  AD,  prove  that  the  triangles  ABC  and 
ADC  are  congruent,  the  line  A  C  bisects  Z  BCD, 
and  EC  equals  CD. 

Ex.  6.    Prove  that  if  a  line  from  a  vertex  of 
a  triangle   perpendicular   to  the   opposite   side  bisects  that  side,  the 
triangle  is  isosceles. 

Ex.  7.    Draw  two  angles  that  are  adjacent  and  not  supplementary; 
adjacent  and  not  complementary. 

Ex.  8.    Of  two  unequal  angles  which  has  the  greater  supplement? 

Ex.  9.    The  complement  of  a  certain  angle  added  to  the  supplement 
of  the  same  angle  is  176°.     Find  the  angle. 

Ex.  10.   What  angle  added  to  one  fifth  of  its  supplement  equals  a  right 
angle  ? 

Ex.  11.   In  the  figure  of  51,  if  Z  A  OM  is  100°,  how  many  degrees  are 
there  in  each  of  the  other  angles  at  0  ? 


TRIANGLES 


17 


\ 


=  to  PB. 


3. 
4. 
5. 

6. 


Construction, 
Identical.  / 


Quote  53. 
Quote  ^7, 


PROPOSITION  III.     THEOREM 

54.  Only  one  perpendicular  can  be 
drawn  to  a  line  from  an  external  point. 

Given:  PR  _L  to  AB  from  P,  and  PD 
any  other  line  from  P  to  AB. 

To  Prove  :  PD  is  not  J_  to  AB  ;  that  is, 
PR  is  the  only-L  from  P  to  AB. 

Proof :  1.    Extend  PR  to  S  making  RS 

2.  Draw  DS. 

3.  In  rt.  A  PDR  and  SDR, 

PR  =  RS. 

4.  DR  =  DR. 

5.  .'.A  PDR  is  congruent  to  A  SDR. 

6.  .-.  ZPD«  =  Z  SDR. 

7.  That  is,  Z  PDR  =  half  of  Z  PDS. 

8.  Now  line  PRS  is  a  st.  line. 

9.  .*.  line  PDS  is  not  a  st.  line. 

10.  /.  Z  PDE,  half  of  Z  PDS,  is  wo£  a  rt.  Z. 

11.  /.  PD  is  not  J_  to  ^1#. 

That  is,  PR  is  the  only  J_  from  P  to  AB. 

Q.E.D. 


The  preceding  form  of  demonstration  will  serve  to  illustrate  an  ex- 
cellent plan  of  writing  the  proofs.  It  will  be  observed  that  the  state- 
ments appear  at  the  left  of  the  page  and  their  reasons  at  the  right.  This 
arrangement  will  be  found  of  great  value  in  the  saving  of  time,  both  for 
the  pupil  who  writes  the  proofs  and  for  the  teacher  who  reads  them. 


8.  Construction, 

9.  Quote  39. 

10.  Qoote  36. 

11.  \Quotef 


Historical  Note.  The  proof  of  the  following  theorem  as  given  in 
the  fifth  proposition  of  Euclid's  "  The  Elements,"  the  most  famous 
geometry  that  was  ever  written,  was  considered  by  the  beginners  as  pre- 
senting great  difficulties.  The  theorem  was  therefore  called  by  the 
ancient  teachers,  the  pans  asinorum,  or  the  bridge  of  the  asses.  Euclid 
discussed  only  magnitudes,  not  their  numerical  measures.  Another  note 
(p.  45)  will  tell  more  of  the  author  of  this  renowned  book. 


18 


BOOK   I.     PLANE   GEOMETRY 


PROPOSITION   IV.     THEOREM 

55.  The  angles  opposite  the  equal  sides 
of  an  isosceles  triangle  are  equal. 

Given  :    A  ABC,  AB  =  AC. 
To  Prove  :    Z  #  =  Z  c. 


B  x  c 

Proof:   Suppose  AX  is  drawn  dividing  Z  BAG  into  two 
equal  A  and  meeting  BC  at  X.     In  the  A  ABX  and  ACX, 

AX  =  AX  (Identical)  . 

AB  =  AC  (Given). 

/.EAX  —  /.GAX  (Const.). 

.'.A  A  EX  is  congruent  to  A  ACX  (52). 

.'.  Z.B  =  Z.c  (27). 

Q.E.D. 

56.    COROLLARY.  An  equilateral  triangle  is  equiangular. 


PROPOSITION  V.     THEOREM 

57.  Two  right  triangles  are  congruent  if  the  hypotenuse  and 
an  adjoining  angle  of  one  are  equal  respectively  to  the  hypote- 
nuse and  an  adjoining  angle  of  the  other. 


A  CD  F 

Given :  Rt.  A  ABC  and  DEF ;    AB  =  DE ;  and  Z  A  =  £  D. 
To  Prove :  A  ABC  is  congruent  to  A  DEF. 

Proof :  Place  A  ABC  upon  A  DEF  so  that  Z  A  coincides 
with  its  equal,  Z  D,  and  AC  falls  along  DF. 

Then  AB  coincides  with  DE  and  point  B  falls  exactly  on 
E  (AB  =  DE). 


TRIANGLES 


19 


Now,  from  point  E,  BC  and  EF  are  both  _L  to  DF 

.'.  BC  coincides  with  EF 
.-.A  ABC  is  congruent  to  A  DEF 


(16). 
(54). 
(26). 

Q.E.D. 


Ex.  1.  In  the  adjoining  diagram,  if  Z 1  =  Z  2,  prove 
the  right  triangles  congruent. 

Ex.  2.    By  use  of  27  prove  PA  =  PC. 

Ex.  3.  Prove  that  the  line,  EM,  from  the  vertex 
of  an  isosceles  triangle,  ABC,  and  perpendicular  to 
the  base,  bisects  the  vertex  angle  and  also  bisects 
the  base. 

(First  prove  the  two  right  &  congruent,  and  then 
use  27.) 

Ex.  4.  Prove  that  the  perpendiculars,  SM  and  TP, 
upon  the  equal  sides  of  an  isosceles  triangle,  RST, 
from  the  opposite  vertices,  S  and  T,  are  equal. 

Two  ways :  (a)  Show  rt.  A  PST  and  MST  are  con- 
gruent; or  (b)  show  rt.  &RSM  and  RPT  are  con- 
gruent. 

Ex.  6.   Prove  that  the  bisector  of  the  vertex  angle  of  an  isosceles  tri- 
angle bisects  the  base  at  right  angles. 

58.  Homologous  parts.     Triangles  are  proved  congruent 
in  order  that  their  homologous  sides  or  homologous  angles 
may  be  proved  equal. 

59.  Auxiliary  lines.     Often  it  is  impossible  to  give  a  simple 
demonstration  without  drawing  lines  not  described  in  the 
hypothesis.     Such  lines,  used  only  for  the  proof,  are  usually 
dotted  in  order  to  distinguish  them  from  the  lines  mentioned 
in  the  hypothesis  and  the  conclusion. 

60.  Elements  of  a  theorem.     Every  theorem  contains  two 
parts.     The   one  is  assumed  to   be  true ;  the  other  results 
from  this  assumption.     The  one  part  contains  the  given  con- 
ditions ;  the  other  part  states  the  resulting  truth. 

The  assumed  part  of  a  theorem  is  called  the  hypothesis. 


20  BOOK   I.     PLANE   GEOMETRY 

The  part  of  the  theorem  which  is  to  be  proved  true  is  the 
conclusion. 

Often  the  hypothesis  is  a  clause  introduced  by  the  word 
"if."  When  this  conjunction  is  omitted,  the  subject  of  the 
sentence  is  known  and  its  qualities,  described  in  the  quali- 
fying words,  constitute  the  "given  conditions."  Thus,  in 
the  theorem  of  52,  the  assumed  part  follows  the  word 
"  if,"  and  the  truth  to  be  proved  is  :  "  Two  triangles  are 
equal." 

The  converse  of  a  theorem  is  the  theorem  obtained  by  inter- 
changing the  hypothesis  and  the  conclusion  of  the  original 
theorem.  Consult  44  and  45,  55  and  114. 

NOTE.  Every  theorem  having  a  simple  hypothesis  and  a  simple  con- 
clusion has  a  converse,  but  only  a  few  of  these  converse  theorems  are  true. 

61.  Elements  of  a  demonstration.     All  correct  demonstra- 
tions should  consist  of  certain  distinct  parts,  namely  : 

1.  Full  statement  of  the  given  conditions  as  applied  to  a 
particular  figure. 

2.  Full  statement  of  the  truth  which  it  is  required  to 
prove. 

3.  The  proof  —  a  series  of  successive  statements,  for  each 
of  which  a  valid  reason  should  be  quoted.     (The  drawing  of 
auxiliary  lines  is  sometimes  essential.) 

4.  The  conclusion  declared  to  be  true. 

PROPOSITION  VI.     THEOREM 

62.  Two  lines  in  the  same  plane  and  perpendicular  to  the 
same  line  are  parallel. 

Given:  CD  and  EF  in  same  plane     c~ 
and  both  _L  to  AB. 

To  Prove :  CD  and  EF  II. 

Proof.  If  CD  and  EF  were  not  II,  they  would  meet  if 
sufficiently  prolonged. 


PARALLEL  LINES  21 

CD  and  EF  are  both  J_  to  AB  (Given). 

.*.  there  would  be  two  lines  _L  to  AB   from   the  point   of 

meeting.     But  this  is  impossible 

.•.CD  and  EF  do  not  meet  and  are  || 

Q.E.D. 

PROPOSITION  VII.     THEOREM 

63.  Two  lines  in  the  same  plane  and  parallel  to  the  same 
line  are  parallel. 

Given :  AB  II  to  BS,  and  CD  II  to  BS,        ^  p 

in  the  same  plane. 

r>  «~ 

To  Prove :  AB  II  to  CD. 

Proof :  If  AB  and  CD  were  not  II,  they  would  meet  if  suf- 
ficiently prolonged. 

AB  and  CD  are  both  II  to  RS  (Given). 

.'.  there  would  be 'two  lines  II  to  BS  through  the  point  of 

meeting.     But  this  is  impossible  (Ax.  13). 

.'.  AB  and  CD  do  not  meet  and  are  ||  (21). 

Q.E.D. 

PROPOSITION  VIII.     THEOREM 

64.  If  a  line  is  perpendicular  to  one  of  two  parallels,  it  is 
perpendicular  to  the  other  also. 

Given:  LM  _L  to  AB  and  AB  II  to       A~~ 
CD.  X 

To  Prove :  LM  ±  to  CD. 
Proof:  Suppose  XY  is  drawn  through  M  _L  to  LM 

XT  is  II  to  AB  9  (62). 

But  CD  is  II  to  AB  (Given). 

And  CD  and  XY  both  contain  point  M          (Const.). 

.*.  CD  and  XY  coincide  (Ax.  13). 

But  LM  is  J_  to  XY  (Const.). 

That  is,  LM  is  ±  to  CD  Q.E.D. 


M Y 


22  BOOK   I.     PLANE   GEOMETRY 

65.    If  one  line  cuts  other  lines,  it  is  called  a  transversal 
Angles  are  formed  at  the  several  intersections,  as  follows  : 

b,  c,  e,  h  are  interior  angles, 
a,  «?,/,  g  are  exterior  angles. 

b  and  A,  c  and  e,  a  and  g,  d  and/ (on  opposite  sides  of 
the  transversal)  are  alternate  angles. 

b  and  h,  c  and  e  are  alternate  interior  angles. 

a  and  g,  d  and  /are  alternate  exterior  angles. 

a  and  e,  d  and  h,  b  and  /  c  and  g  are  corresponding  angles. 


PROPOSITION  IX.     THEOREM 

66.  If  a  transversal  intersects  two  E 

parallels,  the  alternate  interior  angles       R  \\/ 

are  equal.  A~ 

Given:  AB  II  to  CD;  transversal  EF    c. 
cutting  the  Us  at  H  and  K. 

To  Prove :   Z  a  =  Z  i  and  Z  x  =  Z  v. 

Proof:  Suppose  through  3f,  the  midpoint   of   HK,  ES  is 
drawn  J_  to  AB.     Then  £S  is  _L  to  CD.  (64). 

In  rt.  A  EMH  and  KMS,  HM  =  JOf  (Const.). 

.'.A  EMH  is  congruent  to  A  KM8  (57). 

.-.Za  =  Z;  (27). 

Again  Z  ^  is  the  supplement  of  Z  a  (44). 

Also  Z  v  is  the  supplement  of  Z  e  (44). 

.-.Z*  =  Z»  (49). 

Q.E.D. 


Ex.  1.  If  a  line  through  the  vertex  of  an  isosceles 
triangle  is  parallel  to  the  base,  it  makes  equal  angles 
with  the  sides  of  the  triangle. 

Ex.  2.  If  from  each  point  at  which  a  transversal 
intersects  two  parallels  a  perpendicular  to  the  other 
parallel  is  drawn,  two  congruent  right  triangles  are 
formed. 


PARALLEL   LINES 

PROPOSITION  X.     THEOREM 

67.  If  a  transversal  intersects  two 
parallels,  the  corresponding  angles  are 
equal.  A 

Given :  AB  II  to  CD ;  transversal  EF 
cutting  the  Us  and  forming  the  8  A. 

To  Prove  :  Z  s  =  Z  i;  Z  <?  =  Z  r;  Z  0 
=  Z  a;  Z  n  =  Zm. 

Proof:    I.  Zs  =  Z# 


y* 


II.  Z  c  =  Z  m 

Z.m  =  Z.  r 

.'.  Ztf  =  Z  7* 

Etc. 


Ex.  1.  If  a  line  intersects  the  equal  sides  of  an 
isosceles  triangle  and  is  parallel  to  the  third  side, 
the  triangle  formed  has  two  equal  angles. 


Ex.  2.  In  the  adjpining  figure,  if  A  E  is  parallel 
to  EC,  prove  that  Z.  1  +  Z  2  or  Z  CAD  =  Z  3  +  Z4. 

Ex.  3.   In  the  same  figure  prove  that 
Z3+Z4  +  Z5  =  2rt.  A 

Ex.  4.  By  drawing  a  line,  7X/2,  through  the  ver- 
tex A,  of  a  triangle  ^4BC,  parallel  to  EC,  prove 
that  the  sum  of  the  angles  of  any  triangle  is  equal 
to  two  right  angles. 

Ex.  5.  If  the  equal  sides  of  an  isosceles  triangle  are  prolonged  through 
the  vertex,  and  a  line  is  drawn  parallel  to  the  base,  cutting  these  pro- 
longations, this  triangle  formed  has  two  equal  angles. 

Ex.  6.  If  two  sides  of  any  triangle  are  prolonged  beyond  the  third 
side,  and  a  line,  parallel  to  the  third  is  drawn  cutting  these  prolonga- 
tions, two  mutually  equiangular  triangles  are  formed. 


24  BOOK   I.    PLANE   GEOMETRY 

PROPOSITION  XI.     THEOREM 

68.  If  a  transversal  intersects  two  parallels,  the  alternate 
exterior  angles  are  equal. 

Given:  (?)        To  Prove:  (?) 

Proof :  Zc  =  Z  r  (?);  Zr  =  Zw(?).     .'.  Z  c  =  Z.n  (?)  etc. 

Q.E.D. 

PROPOSITION  XII.     THEOREM 

69.  If  a  transversal  intersects  two  S. 


parallels,  the   interior   angles    on   the 

same  sid 

mentary. 


same  side  of  the  transversal  are  supple-    "  7rr*          B 


Given:  (?) 
To  Prove  :  Z  a  +  Z  r  —  2  rt.  A.  etc.          F 
Proof  :  Za  +  Zw  =  2rtZs  (46). 

But  Zw  =  Zr  (66). 

Zr  =  2rt.  Z  etc.     (Ax.  6).   Q.E.D. 


PROPOSITION  XIII.     THEOREM 

70.    If  a  transversal  intersects  two   lines   and  the   alter- 
nate  interior  angles   are   equal,   the  E 

lines  are  parallel.     [Converse  of  66.] 


Given :    AB    and    CD  .two    lines ;  V 


transversal  EF  cutting  them  at  H  and     R ^ 

K,  respectively;  Z  a  —  Z  fTETD.  / 

To  Prove :  CD  II  to  AB. 

Proof :  Through  JT  suppose  ES  is  drawn  II  to  AB. 

Then  Za  =  ZTO<S  (66). 

But  Za  =  Zff.O)  (Hyp.)- 

.'.^HKS=Z.HED  (Ax.  1). 

That  is,  -ffD  and  .HTS  coincide,  and  CD  and  BS  are  the  same 
line.  .-.  CD  is  II  to  AB. 

(Because  it  coincides  with  BS,  which  is  II  to  AB.)      Q.E.D. 


PARALLEL   LINES  25 

PROPOSITION  XIV.     THEOREM 

71.  If  a  transversal  intersects  two  lines  and  the  correspond- 
ing angles  are  equal,  the  lines  are  parallel.   [Converse  of  67.] 

Given :  AB  and  CD  cut  by  EF ;  Z  c  —  Z  r.  (Fig.  in  69.) 
To  Prove :  AB  II  to  CD. 

Proof:                              Z<?=Zra  (51). 

Z,=  Zr  (Hyp.). 

.-.Z™  =  Zr  (Ax.  1). 

.'.  AB  is  II  to  CD.  (70). 

Q.E.D. 

PROPOSITION  XV.     THEOREM 

72.  If  a  transversal  intersects  two  lines  and  the  alternate 
exterior  angles  are  equal,  the  lines  are  parallel.     [Converse 
of  68.] 

Given :  AB  and  CD  cut  by  EF,  and  Z  c  =  Z  n. 
To  Prove :  AB  \\  to  CD. 

Proof:  Zc=Zw  (51). 

Z<?=Zw  (Hyp.). 

.vZw=Zrc  (Ax.  1). 

.-.  .41?  is  II  to  CD  (71).     Q.E.D. 

PROPOSITION  XVI.     THEOREM 

73.  If  a  transversal  intersects  two  lines  and  the  interior 
angles  on  the  same  side  of  the  transversal  are  supplementary, 
the  lines  are  parallel.     [Converse  of  69.] 

Given :  AB  and  CD  cut  by  EF  and  Z  a  +  Z  r  =  2  rt.  A. 
To  Prove :  AB  II  to  CD. 

Proof :              Z  «  is  the  supplement  of  Z  0  (44) . 

Z  a  is  the  supplement  of  Z  r  (Hyp.). 

.-.Z<?=Zr  (49). 

.-.^1B  is  II  to  CD  (71).     Q.E.D. 

ROBBINS'S  NEW  PLANE  GEOM. — 3 


26  BOOK  I.     PLANE   GEOMETRY 

PROPOSITION  XVII.     THEOREM 

74.  If  two  angles  have  their  sides  parallel  each  to  each,  the 
angles  are  equal  or  supplementary. 

NOTE.     There  are  three  cases :    (I)  the  pairs  of  sides  extending  in 
the  same  two  directions  from  the  vertices ;  (II)  the  pairs  of  sides  extend- 
ing in  opposite  directions  from  the  vertices;  (III)  one  pair  extending  in 
the  same  direction  and  the  other  pair  in  op- 
posite directions  from  the  vertices.  / 

I.  Given :    Z  a  and   Z  b,  with  their 
sides  II  each  to  each  and  extending  in 
the  same  directions  from  their  vertices. 

To  Prove :  Z  a  =  Z  £. 

Proof :  If  the  non-parallel  lines  do  not  meet,  produce  them 
to  meet,  forming  Z  0.  Z  a  =  Z  o  (67). 

Z0  =  Z6  (67). 

.-.  Za=Z&  (Ax.  1). 

Q.E.D. 

II.  Given :  Z  a  and   Z  c  with  their  sides  II  each  to  each 
and  extending  in  opposite  directions  from  their  vertices. 

To  Prove :  Z  a  =  Z  c 

Proof :                                 Z  a  =  Z  b  (Proved  in  I). 

Z6  =  Z,  (51). 

.•.Za  =  Z*  (Ax.  1). 

Q.E.D. 

III.  Given :  Z  a  and  Z  d  with  their  sides  II  ;  one  pair  ex- 
tending in  the  same  direction  and  the  other  pair  in  opposite 
directions  from  their  vertices. 

To  Prove :         Z  a  is  supplementary  to  Z  d. 

Proof :  Z  b  is  supplementary  to  Z  c?  (44). 

But  Z  a  =  Z  b  (Proved  in  I). 

Substituting,  Zaissupp.  to  Z  d.  (Ax.  6).  Q.E.D. 

The  proof  that  Z  a  and  /.e  are  supplementary  is  the  same. 


PERPENDICULARS  27 

PROPOSITION  XVIII.     THEOREM 

75.  If  two  angles  have  then*  sides 
perpendicular  each  to  each,  the  angles 
are  equal  or  supplementary. 

I.  Given:  A  a  and  b  with  sides  JL       \j     ^  c 

\&^n 

each  to  each.  jjj^2 — ^ 

To  Prove  :  Z  a  =  Z  b. 

Proof :  At  B  suppose  BR  is  drawn  J_  to  BC  and  BS  _L  to  AB. 

BR  is  II  to  FEa,nd  BSis  II  to  DE  (64). 

.-.Z*=Z5  (74). 

Now  Z  a  is  the  complement  of  Z  y 

Also  Z  #  is  the  complement  of  Z  y 

'    /  a  —  /  r  (\R\ 

.  .  z.  a  —  z_  x  v*0^* 

/.Za  =  Z6  (Ax.  1). 

II.  Given :  A  a  and  c  with  sides  J_  each  to  each. 
To  Prove  :  Z  a  and  Z  0  supplementary. 

Proof :  Z  6  and  Z  c  are  supplementary  (44). 

But  Z  a  =  Z  6  (Proved  in  I). 

Substituting,  Z  a,  and  Z  c  are  supplementary          (Ax.  6). 

Q.E.D. 


A 


Ex.  1.  If  a  line  is  drawn  through  the  vertex  of  an  angle 
and  perpendicular  to  the  bisector  of  the  angle,  it  makes 
equal  angles  with  the  sides. 


,  Ex.  2.  Draw  figure  for  Proposition  XVII  showing  Z.  b  within  Z  a,  and 
prove  the  theorem. 

Ex.  3.  Draw  figure  for  Proposition  XVIII  showing  Z  b  without  Z  a, 
and  prove  the  theorem, 

Ex.  4.  Prove  Proposition  XVIII  if  the  given  angles  have  the  same 
vertex  ;  if  the  vertex  of  one  angle  is  on  a  side  of  the  other. 

Ex.  5.  In  figure  of  75,  if  Za  =  40°,  tell  how  many  degrees  there  are  in 
A  b,  c,  x,  y,  and  SB  C, 


28  BOOK   I.     PLANE   GEOMETEY 

PROPOSITION  XIX.     THEOREM 

76.  Two  triangles  are  congruent  if  a  side  and  the  two 
angles  adjoining  it  in  the  one  are  equal  respectively  to  a  side 
and  the  two  angles  adjoining  it  in  the  other. 

D  L 


B 

Given:  A  BCD  and  JKL  ;  BC=JK 

To  Prove  :  A  BCD  is  congruent  to  A  JKL. 

Proof :  Place  A  BCD  upon  A  JKL  so  that  BC  coincides  with 
its  equal,  JK. 

BD  falls  on  JL  (Because  ^B  is  given  =  to  Z  «7). 

CD  falls  on  KL  (Because  Z  C  is  given  =  to  Z  K). 

Then  point  D,  which  falls  on  both  the  lines  JL  and  ITL,  falls 

at  their  intersection,  L  (38). 

.*.  the  A  are  congruent  (26). 

Q.E.D. 

77.  COROLLARY.  Two  right  triangles  are  congruent  if  a  leg 
and  the  adjoining  acute  angle  of  one  are  equal  respectively  to 
a  leg  and  the  adjoining  acute  angle  of  the  other. 

A B_ 

Ex.  1.  If  AB  is  II  to  DC  and  point  0  bisects 
transversal  BD,  prove  that  it  also  bisects  transver- 
sal AC. 

Ex.  2.   In  the  accompanying  figure,  the  three          ^L i£ 

Us  are  cut  by  two  transversals,  AB  =  BC,  AR  B>f     \          \E 

and  BS  are  II  to  DF.    Prove  that  AR  =  BS.  ~/\  ^^F 


Ex.  3.  In  the  accompanying  figure,  AB 
is  II  to  DC  and  AD  is  II  to  BC.  Prove  that 
the  A  are  congruent. 


/ 

A 


TRIANGLES 


29 


PROPOSITION  XX.     THEOREM 

78.  Two  triangles  are  congruent,  if  the  three  sides  of  one 
are  equal  respectively  to  the  three  sides  of  the  other. 


Given:  &ABC  and  RST;  AB  =  RS-,  AC=RT-,  BC=*ST. 

To  Prove:  A  RST  is  congruent  to  A  ABC. 

Proof:  Place  A  ABC  in  the  position  of  A  AST,  so  that  the 
longest  equal  sides,  BC  and  ST  coincide,  and  A  is  opposite  8T 
from  JR.  Draw  RA. 


Now 


Also 


Also 
Adding, 


RS  =  AS 
'.  AASR  is  an  isosceles  A 

TR=  TA 

'.AATR  is  an  isosceles  A 
.'.Z.SRA  =  /.SAR 
Z  TEA  =  Z  TAR 


(Hyp.)- 
(Def.). 


/-SRT  =  /-SAT 

.'.  A  RST  is  congruent  to  A  AST 
That  is,  by  substituting,  A  RST  is  congruent  to  A  ABC  (Ax.  6) . 

Q.E.D 


(Def.). 
(55). 
(55). 
(Ax.  2). 
(52). 


Ex.     In  the  figure  of  Ex.  3  on  the  opposite  page,  if  the  opposite  sides 
are  equal,  prove  them  parallel. 

79.    The  distance  from  one  point  to  another  is  the  length 
of  the  straight  line  joining  the  two  points. 


30 


BOOK   I.     PLANE   GEOMETRY 


PROPOSITION  XXI.     THEOREM 

80.  Any  point  in  the  perpendicular 
bisector   of  a  line   is  equally  distant 
from  the  extremities  of  the  line. 

Given :  AE  J_  to  CD  at  its  midpoint, 
B  ;  P  any  point  in  AE  ;  PC  and  PD. 
To  Prove :  PC  =  PD. 
Proof :  In  the  rt.  A  PBC  and  PBD, 
PB  =  PB 

BC  =  BD 

.'.A  PBC  is  congruent  to  A  PBD 
.'.  PC  =  PD. 

PROPOSITION  XXII.     THEOREM 

81.  Any  point  not  in  the   perpen- 
dicular bisector  of  a  line  is  not  equally 
distant  from  the  extremities  of  the  line. 

Given :  AB  J_  bisector  of  CD ;  P  any 
point  not  in  AB  ;  PC  and  PD. 
To  Prove :  PC  =£  PD. 


(Identical). 

(Hyp.). 

(53). 

(27). 

Q.E.D. 


Proof:  Either  PC  or  PD  will  cut  AB. 
AB  at  O.     Draw  OD. 

DO  +  OP  >  PD 

But  CO  =  DO 

Substituting,  CO  4-  OP  >  PD 

That  is,  PC  >  PD  or  PC  =£  PD. 


Suppose  PC  cuts 

(Ax.  12). 

(80). 

(Ax.  6). 

Q.E.D. 


82.  COROLLARY.     If  a  point  is  equally  distant  from  the  ex- 
tremities of  a  line,  it  is  in  the  perpendicular  bisector  of  the 
line.  (80  and  81.) 

83.  COROLLARY.     Two  points  each  equally  distant  from  the 
extremities  of  a  line  determine  the  perpendicular  bisector  of 
the  line.  (82  and  4.) 


TRIANGLES 


31 


PROPOSITION  XXIII.     THEOREM 

84.  Two  right  triangles  are  equal  if  the  hypotenuse  and  a 
leg  of  one  are  equal  respectively  to  the  hypotenuse  and  a  leg 
of  the  other. 

K  R 


Given :  rt.  A  UK  and  LMR  ;  Kl  =  RM  ;  KJ  =  RL. 
To  Prove :  A  UK  =  A  LMR. 

Proof :  Place  A  UK  in  the  position  of  A  jr£jR  so  that  the 
equal  sides,  KJ  and  RL,  coincide,  and  I  is  at  X,  opposite  RL 
from  M. 

A  RLM  and  RLX  are  supplementary 

.*.  -ZX.M"  is  a  straight  line 

.'.  figure  XMR  is  a  A 

RX  =  R  M 

.'.  A  XMR  is  isosceles 


Now 


Now 


(19). 
(45). 
(23). 

(Hyp.)- 
(Def.). 
(55). 
(57). 
That  is,  A  UK  is  congruent  to  A  LMR  (Ax.  6). 

Q.E.D. 

85.  COROLLARY.  The  perpendicular  from  the  vertex  of  an 
isosceles  triangle  to  the  base  bisects  the  base,  and  bisects  the 
vertex  angle. 

In  the  congruent  right  A  of  84,  XL  =  LM  (27). 

Also  Z.  XRL  =  MRL  (27). 

Q.E.D. 


.A  XLR  is  congruent  to  A  LMR 
A  UK  is  congruent  to  A  LMR 


Ex.     In  the  figure  of  84,  if  XL  —  LM,  prove  by  two  methods  that 
XR  =  RM. 


32 


BOOK   I.     PLANE   GEOMETRY 


PROPOSITION  XXIV.     THEOREM 

86.  The  sum  of  two  sides  of  a  triangle  is  greater  than  the 
sum  of  two  lines  drawn  to  the  ex- 
tremities of  the  third  side,  from  any 
point  within  the  triangle. 

Given:   P,  any  point  in  A  ABC; 
lines  PA  and  PC. 

To  Prove :  AB  +  BC>  AP  +  PC. 

Proof :  Extend  AP  to  meet  BC  at  X. 


IU^ 
Subtract 


AB  +  BX  >  AP  +  PX  (Ax.  12). 

CX  4-  PX  >  PC  (Ax.  12). 


+  BX  +  CX  +  PX  >  AP  +  PC  +  PX      (Ax.  8). 
PX  =  PX 

.'.  AB  H-  BC>  AP  +  PC       (Ax.  7).  Q.E.D. 


Ex.   If  from  any  point  within  a  triangle  lines 
are  drawn  to  the  three  vertices  : 

(1)  their  sum  is  less  than  the  sum  of  the  sides 
of  the  triangle. 

(2)  their  sum  is  greater  than  half  the  sum  of 
the  sides  of  the  triangle. 

PROPOSITION  XXV.     THEOREM 

87.  The  perpendicular  is  the  shortest 

line  that  can  be  drawn  from  a  point  to  a    A 

straight  line. 

Given :  PR  -L  to  AB  ;  PC  not  _L. 

To  Prove :  PR  <  PC. 

Proof :  Extend  PR  to  X,  making  RX  =  to  PR. 
(1)  PR  +  RX  <  PC  +  ex 

But  AR  is  -L  bisector  of  PX 

.-.CX=  PC 
Also  RX  =  PR 


X 

Draw  CX. 

(Ax.  12). 

(Const.). 

(80). 

(Const.). 


TRIANGLES 


33 


.-.  Substituting  in  (1),  PR  +  PB  <PC  +  PC  (Ax.  6). 

That  is,  2  PR  <  2  PC. 

.'.  PU  <  PC  (Ax.  10).       Q.E.D. 


PROPOSITION  XXVI.     THEOREM 

88.  If  from  any  point  in  a  perpendicular  to  a  line  two  oblique 
lines  are  drawn, 

I.  Oblique  lines  cutting  off  equal  distances  from  the  foot 
of  the  perpendicular  are  equal. 

n.   Equal  oblique  lines  cut  off  equal  distances.    [Converse.] 

III.  Oblique  lines   cutting    off  unequal  distances  are  un- 
equal, and  that  one  which  cuts  off  the  greater  distance  is  the 
greater. 

IV.  Unequal  oblique  lines  cut  off  unequal  distances  from 
the  foot  of   the   perpendicular,   and   the 

longer  oblique  line  cuts  off  the  greater 
distance.  [Converse.] 

I.  Given :  CD  -L  to  °AB  ;  ND  =  MD  ; 
oblique  lines  PN  and  PM. 

To  Prove :  PAT  =  PM. 


N 


M 


Proof :  In  the  rt.  A  PDN  and  PDM,  PD  =  PD 
Also  ND  =  DM 

.'.  A  PDN  is  congruent  to  A  PDM 
.'.  PN  =  PM 

II.    Given :  CD  _L  to  AB  ;  PN  =  PM. 
To  Prove :  DN  =  DM. 

Proof :  In  the  rt.  A  PDN  and  PDM,  PD  =  PD 
Also  PN  =  PM 

.'.  A  PDN  is  congruent  to  A  PDM 
.'.  DN=  DM 


(Iden.). 

(Hyp.)- 
(53). 

(27). 

Q.E.D. 


(Iden.). 

(Hyp.)- 
(84). 

(27). 

Q.E.D. 


34  BOOK   I.     PLANE   GEOMETRY 


III.  Given :    CD  _L  to   AB  ;    oblique 
lines  PE  and  PT  : 

Also  ED  >  DT.  A 

To  Prove :        PR>PT. 


Proof :    Because  DR  >  DT,  we  may  take  DS  (on  XXR)  =  to 

DT.    Draw  PS.     Extend  PD  to  X,  making  DX  =  to  PD.     Draw 

EX  and  SX.     Now  AD  is  the  J_  bisector  of  PX  and  CD  is  the 

_L  bisector  of  ST  (Const.). 

PE  +  RX>PS+  sx  (86). 

But  EX  =  PR,  and  SX  =  PS  =  PT  (80). 

Substituting,          PR  -h  Pfl  >  PS  +  PS  (Ax.  6). 

That  is,  2  PR  >  2  PS 

.'.  PR>PS  (Ax.  10). 

Substituting,  PR  >  PT  (Ax.  6).     Q.E.D. 

IV.    Given:    [Use  no  dotted  lines.]    CD_Lto  AB  ;   oblique 
lines  PR  and  PT ;  P#  >  PT. 

To  Prove :  DR  >  DT. 

Proof :  It  is  evident  that  DR  <  DT,  or  DR  —  DT,  or  DR  >  DT. 

IST:  If  DR  <  Dr,  PE  <  PT  (By  III). 

But  PR>PT  (Hyp.). 

.'.  DR  is  not<DT 

2D:  If  DjR  =  DT,  PR=PT  (By  I). 

But  PR  >  PT  (Hyp.). 

.'.  DR  is  not  =  DT 
3D:  Therefore,  the  only  possibility  is  that  DR  >  DT. 

Q.E.D. 

89.   COROLLARY.   From  an  external  point  it  is  not  possible  to 
draw  three  equal  straight  lines  to  a  given  straight  line. 


TRIANGLES  35 

90.  The  method  of  demonstration  employed  in  88,  IV,  is 
called  the  method  of  exclusion. 

It  consists  in  making  all  possible  suppositions,  leaving  the 
probable  one  last,  and  then  proving  all  these  suppositions 
impossible,  except  the  last,  which  must  necessarily  be  true. 

The  method  of  proving  the  individual  steps  is  called  re- 
ductio  ad  absurdum  (reduction  to  an  absurd  or  impossible 
conclusion).  This  method  consists  in  assuming  as  false  the 
truth  to  be  proved  and  then  showing  that  this  assumption 
leads  to  a  conclusion  altogether  contrary  to  known  truth  or 
the  given  hypothesis.  (Examine  the  last  preceding  proof.) 
This  is  sometimes  called  the  indirect  method.  The  theorems 
of  62  and  63  are  demonstrated  by  a  single  use  of  this  method. 

PROPOSITION  XXVII.     THEOREM 

91.  If  two  triangles  have  two  sides  of  one  equal  to  two 
sides  of  the  other,  but  the  included  angle  in  the  first  greater 
than  the  included  angle  in  the  second,  the  third  side  of  the 
first  is  greater  than  the  third  side  of  the  second. 

Given:   A  ABC,  DEF; 

AB  =  DE',    BC=EF; 
Z  ABC  >  Z.E. 

To  Prove :  AC>  DF. 

Proof :  Place  the  A  DEF  upon  A  ABC  so  that  side  DE  coin- 
cides with  its  equal  AB,  A  DEF  taking  the  position  of  A  ABH. 
There  remains  an  /.  HBC.  (Hyp.). 

Draw  HC  and  suppose  its  JL  bisector,  MX,  to  be  erected, 
meeting  AC  at  X.  Draw  HX. 

Now  HX  =  XC  (80). 

Also  AX  +  xn  >  AH  (Ax.  12). 

Substituting,        AX  +  XC,  or  AC  >  AH  (Ax.  6). 

Substituting,  AC>  DF  (Ax.  6).     Q.E.D. 


36  BOOK  I.     PLANE   GEOMETRY 

PROPOSITION  XXVIII.     THEOREM 

92.  If  two  triangles  have  two  sides  of  one  equal  to  two  sides 
of  the  other,  but  the  third  side  of  the  first  greater  than  the 
third  side  of  the  second,  the  included  angle  of  the  first  is 
greater  than  the  included 

r> 

angle  of  the  second.  [Con- 
verse.] 

Given :  A  ABC  and  RST ; 

AB  —  RS  ;   BC  =  8T  ;  A* ^C  R4 

AC  >  RT. 

To  Prove :  Z  B  >  Z  s. 

Proof :  Z.B  <  Z  s,  or  Z  B  =  Z  s,  or  Z  B  >  Zs. 

1.  IfZ.B<Z/S,  AC  <  RT  (91). 
But                                   AC  >  RT                               (Hyp.). 

.*.  Z  B  is  not  <  Z  s. 

2.  If  Z  B  =  Z  5,  the  A  are  congruent  (52). 

•      AT—   -PT  ff)rl\ 

.  .  AL  —  Kl  (/•> 

But  AC  >  RT  (Hyp.). 

3.  .*.  the  only  possibility  is  that  Z  B  >  Z  s.  Q.E.D. 

93.  The  distance  from  a  point  to  a  line  is  the  length  of 
the  perpendicular  from  the  point  to  the  line.     If  the  per- 
pendiculars from  a  point  to  two  lines  are  equal,  the  point  is 
said  to  be  equally  distant  from  the  lines. 

PROPOSITION  XXIX.    THEOREM 

94.  Every  point  in  the  bisector  of  an 
angle  is  equally  distant  from  the  sides 
of  the  angle. 

Given:   /.ACE;    bisector  CQ ;    point 
P  in  CQ;  distances  PB  and  PD. 
To  Prove :  PB  =  PD. 


TRIANGLES  37 

Proof:    A  PEC  and  PDC  are   rt.   A. 

In  rt.  A  PBC  and  PDC,  PC  =  PC  (Iden.). 

Z.PCB  =  ^PCD  (Hyp.). 

/.A  PBC  is  congruent  to  A  PDC  (57). 

PB  =  PD  (27).       Q.E.D. 

PROPOSITION  XXX.     THEOREM 

95.  Every  point  equally  distant  from  the  sides  of  an  angle 
is  in  the  bisector  of  the  angle. 

Given :  Z.  A  CE ;  P,  a  point,  such  that  PB  =  PD  (distances)  ; 
CQ,  a  line  from  vertex  of  the  angle,  and  containing  P. 
To  Prove :  Z  ACQ  =  Z  ECQ. 

Proof :  A  P£C  and  PDC  are  rt.  A  (93). 

In  rt.  A  PBC  and  PDC,  PC  =  PC  (Iden.). 

PB  =  PD  (Hyp.) 

.*.  A  PBC  is  congruent  to  A  PDC  (84). 

.' .  Z.  ACQ  = /.  ECQ  (27).      Q.E.D. 

96.  COROLLARY.     If  a  point  is  not  equally  distant  from  the 
sides  of  an  angle,  it  is  not  in  the  bisector  of  the  angle. 

(If  it  were  in  the  bisector,  it  would  be  equally  distant.) 

97.  COROLLARY.    The  vertex  of  an  angle  and  a  point  equally 
distant  from  its  sides  determine  the  bisector  of  the  angle. 

98.  The  altitude  of  a  triangle  is  the  perpendicular  from 
any  vertex  to  the  opposite  side  (prolonged  if  necessary).     A 
triangle  has  three  altitudes. 

The  bisecljgj  of  an  angle  of  a  triangle  is  the  line  dividing 
any  angle  into'  two  equal  angles.  A  triangle  has  three  bisec- 
tors of  its  angles. 

The  median  of  a  triangle  is  the  line 
drawn  from  any  vertex  to  the  midpoint  of 
the  opposite  side.  A  triangle  has  three 

medians.  THE  THREE  MEDIANS 


38  BOOK   I.     PLANE   GEOMETRY 

PROPOSITION  XXXI.     THEOREM 

99.  The  bisectors  of  the  angles  of  a  triangle  meet  in  a 
point  which  is  equally  distant  from  the  sides. 

Given :  A  ABC,  AX  bisecting  /.  A,  BY 
and  CZ  the  other  bisectors. 

To  Prove  :  AX,  BY,  CZ  meet  in  a  point 

equally  distant  from  AB,  AC,  and  BC. 

B  C 

Proof :  Suppose  that  AX  and  B  T  intersect  at  O. 
O,  in  AX,  is  equally  distant  from  AB  and  AC  (94). 
O,  in  BY,  is  equally  distant  from  AB  and  BC  (?). 
.'.  point  O  is  equally  distant  from  AC  and  BC  (Ax.  1). 
/.  O  is  in  bisector  CZ  (95). 
That  is,  all  three  bisectors  meet  at  O,  and  O  is  equally  dis- 
tant from  the  three  sides.  Q.E.D. 


PROPOSITION  XXXII.     THEOREM 

100.  The  three  perpendicular  bisec- 
tors of  the  sides  of  a  triangle  meet  in 
a  point  which  is  equally  distant  from 
the  vertices. 

Given:  A  ABC-,  LB,  MS,  NT,  the 
three  J_  bisectors. 

To  Prove:  LE,  MS,  NT  meet  in  a 
point  equally  distant  from  A,  B,  C.  N  C 

Proof :  Suppose  that  LB  and  MS  intersect  at  O. 

O,  in  LB,  is  equally  distant  from  A  and  B  (80). 

O,  in  MS,  is  equally  distant  from  A  and  C  (?). 

/.  point  O  is  equally  distant  from  B  and  C    (Ax.  1). 

.'.  O  is  in  _L  bisector  NT  (82). 

That  is,  all  three  _L  bisectors  meet  at  O,  and  O  is  equally 

distant  from  A  and  B  and  C.  Q.E.D. 


TRIANGLES 


39 


101.  An  exterior  angle  of  a  triangle  is  an  angle  formed 
outside  the  triangle,  between    one  side  of 

the   triangle  and   another   side   prolonged. 
[Z.ABX.~] 

The    angles  within  the  triangle  at  the 
other    vertices    are    the    opposite   interior    ^ 
angles.      [Z.4  and  Zc.] 

PROPOSITION  XXXIII.     THEOREM 

102.  An  exterior  angle  of  a  triangle  is  D 
equal  to  the  sum  of  the  opposite  interior 

angles. 

Given :  A  ABC  ;  exterior  Z  ABD. 

To  Prove :  Z  ABD  =  Z  A  +  Z  c. 

Proof :    Suppose  BP  to  be  drawn  through  B  II  to  A  C. 

Z  ABD  =  Z  ABP  +  Z  PBD  (Ax.  4). 

But  ZABP=Z.4  (66). 

Also  Z  PBD  =  ZC  (67). 


=  Z.A  +  ZC 


(Ax.   6).      Q.E.D, 


103.   COROLLARY.     An  exterior  angle  of  a  triangle  is  greater 
than  either  of  the  opposite  interior  angles.  (Ax.  5.) 


Ex.  1.  If  lines  are  drawn  from  any  point  within  a 
triangle  to  two  vertices  of  the  triangle,  they  in- 
clude an  angle  greater  than  the  third  angle  of  the 
triangle. 

[Notice  that  Z  1  is  an  ext.  Z  of  A  CDP  and  Z  2  is    B- 
an  ext.  Z  of  &ABD.~] 

Ex.   2.   If  the  side  LM,  of  equilateral  triangle  LMN, 
is  produced  to  P,  and  PN  is  drawn,  /.P</.L. 


40  BOOK   I.    PLANE   GEOMETRY 


Ex.  3.  The  bisectors  of  two  exterior  angles  of  a 
triangle  and  of  the  interior  angle  at  the  third  ver- 
tex meet  in  a  point. 


Ex.  4.   The  line  through  the  vertex  of  an  isosceles 
triangle,  parallel  to  the  base,  bisects  the  exterior  angle. 

Ex.  6.    The  bisector  of  the  exterior  angle  at  the  ver- 
tex of  an  isosceles  triangle  is  parallel  to  the  base. 

/ 
Proof :  ^  DCB  =  2ZA  (?)  =  2 Z  DCR  (?). 

PROPOSITION  XXXIV.     THEOREM 

104.  The  sum  of  the  angles  of  any 
triangle  is  two  right  angles;  that  is, 
1 80°. 

Given:  A  ABC. 
To  Prove: 

Z  A  +  Z  B  +  Z  ACB  =  2  rt.  A  =  180°. 
Proof:  Prolong  AC  to  X,  making  the  ext.  Z  BCX. 

Z  BCX  +  ^ACB  =  2  rt.  A  (46). 

But  z  BCX  =/.A  +  £B  (102). 

.-.  Z  A  +  Z  B  +  ZACB  =  2  rt.  A  =  180°     (Ax.  ft). 

Q.E.D. 

105.  COROLLARY.     The  sum  of  any  two  angles  of  a  triangle 
is  less  than  two  right  angles.  (Ax.  5.) 

106.  C OROLLARY.     A  triangle  cannot  have  more  than  one  right 
angle  or  more  than  one  obtuse  angle. 

107.  COROLLARY.     Two  angles  of  every  triangle  are  acute. 

108.  COROLLARY.     The  acute  angles  of  a  right  triangle  are 
complementary. 


TRIANGLES  41 

Proof:  Their  sum  =  1  rt.  Z.  (104.) 

Hence  they  are  complementary. 


109.  COROLLARY.    Each  angle  of  an  equiangular  triangle  is 
60°. 

110.  COROLLARY.    If  two  right  triangles  have  an  acute  angle 
of  one  equal  to  an  acute  angle  of  the  other,  the  remaining  acute 
angles  are  equal.  (48.) 

111.  C  OROLLARY.     If  two  triangles  have  two  angles  of  the  one 
equal  to  two  angles  of  the  other,  the  third  angle  of  the  first  is 
equal  to  the  third  angle  of  the  second.  (104.) 

112.  COROLLARY.     Two  triangles  are  congruent  if  a  side  and 
any  two  angles  of  the  one  are  equal  respectively  to  a  homolo- 
gous side  and  the  two  homologous  angles  of  the  other. 

Proof:  The  third  Z  of  one  A  =  third  Z  of  other  A  (111). 
/.the  A  are^  (76). 

113.  COROLLARY.    Two  right  triangles  are  congruent  if  a  leg 
and  the  opposite  acute  angle  of  one  are  equal  respectively  to  a 
leg  and  the  opposite  acute  angle  of  the  other. 


Ex.  1.   If  two  angles  of  a  triangle  contain  50°  and  100°,  how  many 
degrees  are  there  in  the  other  angle  ? 

Ex.  2.    Prove  that  if  one  angle  of  a  triangle  equals  the  sum  of  the 
other  two  angles,  the  triangle  is  a  right  triangle. 

Ex.  3.    How  many  degrees  are  there  in  each  angle  of  an   isosceles 
right  triangle? 

Ex.  4.  The  altitude  of  an  isosceles  right  triangle 
upon  the  hypotenuse  divides  the  triangle  into  two 
isosceles  right  triangles,  each  of  whose  acute  angles 
is  equal  to  45°.  A*~"  fjf 

Ex  6.   In  a  right  triangle,  if  one  acute  angle  is  47°,  what  is  the  other? 
Ex.  6.   In  an  isosceles  triangle,  if  Z  A  =  Z  B  =  80°,  find  Z.  C. 

ROBBINS'S    NEW    PLANE    GEOM. 4 


42 


BOOK   I.     PLANE   GEOMETRY 


it 


=  25°,  Z  B  =  88°,  find  Z  C  and  the  exterior 


Find  each  base 


Ex.  7.   In  A  ABC, 
angle  at  A. 

Ex.  8.   The  vertex  angle  of  an  isosceles  triangle  is  44°. 
angle. 

Ex.  9.  If  one  acute  angle  of  a  right  triangle  is  double  the  other, 
how  many  degrees  are  there  in  each? 

Ex.  10.  If  one  acute  angle  of  a  right  triangle  is  five  times  the  other, 
how  many  degrees  are  there  in  each  ? 

Ex.  11.  If  any  angle  of  an  isosceles  triangle  is  60°,  show  that  the  triangle 
is  equiangular. 

Ex.  12.  If  the  vertex  angle  of  an  isosceles  triangle  equals  four  times 
the  sum  of  the  base  angles,  find  each  angle. 

Ex.  13.  If  the  vertex  angle  of  an  isosceles  tri- 
angle is  twice  the  sum  of  the  base  angles,  any  line 
perpendicular  to  the  base  forms  with  the  sides  of 
the  given  triangle  (one  side  to  be  produced)  an 
equiangular  triangle.  [First  find  Z  x  and  Z  ACB. 
Then  use  exterior  A  at  /).] 

Ex.  14.   The  angles  of  a  triangle  are  44°,  62°,  74 
by  lines  meeting  at  a  point. 
angles  at  this  point. 

Ex.  15.  If  two  angles  of  a  triangle  are  80°  and  55°,  how  many  degrees 
are  there  in  the  angle  formed  by  their  bisectors  ? 

Ex.  16.  The  vertex  angle  of  an  isosceles  triangle  is  one  third  of  either 
exterior  .angle  at  the  extremities  of  the  base.  Find  each  angle  of  the 
triangle. 

Ex.  17.   If  two  angles  of  a  triangle  are  30°  and 
40°,  how  many  degrees  are  there  in  the  angle  formed 
by  the  bisector  of  the  third  angle  and  the  altitude 
from  the  same  vertex?     Solution: 
=  \ZABC  -  comp.  of  Z.A. 


O      B 


These  are  bisected 
Find  the  number  of  degrees  in  the  three 


Ex.  18.  The  angle  between  the  altitude  of  a  triangle  and  the  bisector 
of  the  angle  at  the  same  vertex  equals  half  the  difference  of  the  other 
angles  of  the  triangle. 

Proof.  Zx  =  ABS-ZABD 

=  i  (180°  -  Z  A  -  Z  C)  -  (90°  -  £  A)  (Ax.  6). 

=  etc. 


EXERCISES  43 

Ex.  19.  The  exterior  angle  at  the  base  of  an  isos- 
celes triangle  equals  half  the  vertex  angle  plus  90°. 

Ex.  20.  If  in  A.45C,  ZBAC=80°,  Z  ABC  =  30°, 
find  the  angle  formed  by  the  bisectors  of  the  exte- 
rior angles  at  A  and  B. 

Ex.  21.  The  angle  formed  by  the  bisectors  of 
two  exterior  angles  of  a  triangle  equals  half  the 
sum  of  the  interior  angles  at  the  same  vertices. 

Ex.  22.  If,  in  triangle  ABC,  the  bisectors  of  the 
interior  angle  at  B  and  the  exterior  angle  at  C 
meet  at  D,  the  angle  BAC  equals  twice  the  angle 
BDC.  b 

Ex.  23.    The  angle  between  the  bisectors  of  two  angles  of  a  triangle 
equals  half  the  third  angle,  plus  a  right  angle. 

s 

Ex.  24.  If  from  any  point  in  the  base  of  an  isos- 
celes triangle  perpendiculars  to  the  equal  sides  are 
drawn,  they  make  equal  angles  with  the  base. 


R 

Ex.  26.  If  one  of  the  legs  of  an  isosceles  triangle  is 
produced  through  the  vertex  its  own  length,  and  the  ex- 
tremity is  joined  to  the  nearer  end  of  the  base,  this  line 
is  perpendicular  to  the  base. 

Ex.  26.  If  the  middle  point  of  one  side  of  a  triangle 
is  equally  distant  from  the  three  vertices,  the  triangle 
is  a  right  triangle. 


Ex.  27.  If  the  points  at  which  the  bisectors  of  the  equal  angles  of  an 
isosceles  triangle  meet  the  opposite  sides,  are  joined  by  a  line,  it  is  par- 
allel to  the  base.  . 

A/ 

Ex.  28.   The  bisectors  of  two  interior  angles  on  the 
same  side  of  a  transversal  cutting  two  parallels  meet 


at  right  angles.  — *f*- D 


Proof:                      ZBAC+^ACD  =  180°  (?) 

.-.  $ZBAC  +  $£ACD  =  W°  (Ax.  3). 

That  is,                  Z  MA  C  +  /.  MCA  =  90°  (Ax.  6). 

.-.  ZM=90°  (104), 


44 


BOOK  1.    PLANE  GEOMETRY 


PROPOSITION  XXXV.     THEOREM 

114.  If  two  angles  of  a  triangle  are 
equal,  the  triangle  is  isosceles.  [Con- 
verse of  55.] 

Given :  A  ABC  ;  ^  A  =  Z  c. 

To  Prove :  AB  =  BC. 

Proof :  Suppose  BX  drawn  _L  to  AC. 
In  the  rt.  A  ABX  and  CBX,  BX  =  BX 


*.  A  ABX  is  congruent  to  A  CBX 


(Iden.). 
(Hyp.). 


(27). 
Q.E.D. 


115.  COROLLARY.    An  equiangular  triangle  is  equilateral. 


Ex.  1.   If  the  exterior  angles  of  a  triangle  are 
equal,  the  triangle  is  isosceles. 


Ex.  2.  A  line  parallel  to  the  base  of  an  isosceles 
triangle,  meeting  the  equal  sides,  forms  another  isos- 
celes triangle.  Prove  this  for  all  three  cases,  whether 
it  meets  the  equal  sides,  or  those  sides  prolonged. 

Ex.  3.  If  a  line  through  the  vertex  of  a  triangle 
bisects  the  exterior  angle  and  is  parallel  to  the  base, 
the  triangle  is  isosceles. 


Ex.  4.  If  from  any  point  in  the  bisector  of  an 
angle  a  line  is  drawn  parallel  to  either  side  of  the 
angle,  an  isosceles  triangle  is  formed. 


Ex.  6.   The  bisectors  of  the  equal  angles  of  an  isosceles  triangle  form, 
with  the  base,  another  isosceles  triangle. 


EXERCISES 


45 


Ex.  6.  If  from  any  point  in  the  base  of  an  isosceles  triangle  a  line  is 
drawn  parallel  to  one  of  the  equal  sides  and  meeting  the  other  side,  an 
isosceles  triangle  is  formed. 

A 


Ex.  7.   If  two  altitudes  of  a  triangle  are  equal,  the 
triangle  is  isosceles.     (Prove  two  ways.) 


Ex.  8.  If  a  perpendicular  is  erected  at  any  point  in 
the  base  of  an  isosceles  triangle,  meeting  one  leg,  and 
the  other  leg  produced,  another  isosceles  triangle  is 
i'ormed. 


Ex.  9.  If  ABC  is  a  triangle,  BS  is  the  bisector 
of  ^.ABC,  and  AM  is  parallel  to  BS,  meeting  EC 
produced,  at  M,  the  triangle  ABM  is  isosceles. 


Ex.  10.  If  a  line  is  drawn  perpendicular 
to  the  bisector  of  an  angle  and  intersecting 
the  sides,  an  isosceles  triangle  is  formed. 


Ex.  11.   If  ABC  is  a  triangle  and  BS  is  the  bisector  of  exterior  Z.  ABR 
and  AM  is  II  to  BS,  meeting  BC  at  M,  A  ABM  is  isosceles. 

Historical  Note.  Euclid  of  Alexandria  lived 
during  the  third  century  B.C.  ;  but  little  is  known 
of  his  parentage,  teachers,  residence,  or  career.  He 
was  probably  a  contemporary  of  Eratosthenes  and 
Archimedes.  He  wrote  "  The  Elements,"  the  most 
complete  treatise  on  geometry  that  appeared  before 
modern  times,  and  inasmuch  as  this  supplanted  all 
others  he  must  have  made  a  great  advance  over  the 
work  of  his  predecessors.  He  was  both  a.  compiler  and  a  discoverer. 
When  a  king  of  Egypt  asked  him  about  the  possibility  of  mastering 
geometry  with  ease,  Euclid  is  said  to  have  replied,  "  There  is  no  royal 
road  to  geometry." 


EUCLID 


46  BOOK   I.     PLANE   GEOMETRY 

PROPOSITION  XXXVI.     THEOREM 

116.  If  two  sides  of  a  triangle  are  un- 
equal, the  angle  opposite  the  longer  side 
is  greater  than  the   angle  opposite  the 
shorter  side. 

Given :  A  ABC  ;  AB  >  AC.  c 

To  Prove :  Z  ACB  >  Z.B. 

Proof :  On  AB  take  AR  =  to  AC. 

Draw  CR  and  let  Z  ARC  =  x. 

Z.  ARC  is  an  ext.  Z  of  ACBR  (Def.  101). 

.vZa?>  z#  (103). 

But                             Z.ACR  =  ^X  (55). 

Substituting,             Z.ACR  >  Z#  (Ax.  6). 

Again,                         Z  ACB  >  Z  ACR  (Ax.  5). 

.•.Z^>Z*  (Ax.  11). 

Q.E.D. 

PROPOSITION  XXXVII.     THEOREM 

117.  If  two  angles  of  a  triangle  are 
unequal,  the  side  opposite  the  greater 
angle  is  longer  than  the  side  opposite 
the  less  angle.     [Converse.] 

Given :  A  ABC  ;  Z  ACB  >  Z  B. 
To  Prove:  AB  >  AC. 


Proof:  In  Z  ACB,  suppose  Z  BCR  constructed  =  to 

Then  CR  =  BR                             (H4). 

Also  AR  +  CR  >  AC                      (Ax.  12). 

Substituting,  AR  +  BR  >  AC                       (Ax.  6). 

That  is,  AB  >  AC.                             Q.E.D. 

118.   COROLLARY.  The  hypotenuse  is  the  longest  side  of  a 

right  triangle.  (117.) 


QUADRILATERALS 


47 


QUADRILATERALS 

119.  A  quadrilateral  is  a  portion  of  a  plane  bounded  by 
four  straight  lines.     These  four  lines  are  called  the  sides. 
The  vertices  of  a  quadrilateral  are  the  four  points  at  which 
the  sides  intersect.     The  angles  of  a  quadrilateral  are  the 
four  angles  at  the  vertices.     The  diagonal  of  a  rectilinear 
figure  is  a  line  joining  two  vertices,  not  in  the  same  side. 

120.  A  trapezium  is  a  quadrilateral  having  no  two  sides 
parallel.     A  trapezoid  is  a  quadrilateral  having  two  and  only 
two  sides  parallel.     A  parallelogram  is  a  quadrilateral  having 
its  opposite  sides  parallel  (O). 

NOTE.    In  the  first  figure  below,  angles  A,  Z>,  or  B,  C  are  opposite 
angles,  angles  A,  C,  or  B,  Z>,  or  A,  B,  or  (7,  D  are  consecutive  angles. 


C  D 

PARALLELOGRAM 
RHOMBOID 


SQUARE 


RECTANGLE 


TRAPEZOID 


121.  A  rectangle  is  a  parallelogram  whose  angles  ar_e  right 
angles.  A  rhomboid  is  a  parallelogram  whose  angles  are  not 
right  angles. 


122.    A  square  is  an  equilateral  rectangle. 
an  equilateral  rhomboid. 


A  rhombus  is 


123.  The  side  upon  which  a  figure  appears  to  stand  is 
called  its  base.  A  trapezoid  and  all  parallelograms  have  two 
bases,  —  the  actual  base  and  the  side  parallel  to  it.  The  non- 
parallel  sides  of  a  trapezoid  are  sometimes  called  the  legs.  An 
isosceles  trapezoid  is  a  trapezoid  whose  legs  are  equal.  The 
median  of  a  trapezoid  is  the  line  connecting  the  midpoints  of 
the  legs.  The  altitude  of  a  trapezoid  and  of  all  parallelograms 
is  the  perpendicular  distance  between  the  bases. 


48 


BOOK   I.     PLANE   GEOMETRY 


PROPOSITION  XXXVIII. 

124.  The  opposite  sides  of  a  paral- 
lelogram are  equal. 

Given  :  O  LMOP. 

To  Prove  :  LM  =  PO  and  LP  =  MO. 

Proof  :  Draw  diagonal  PM. 
In  A  LMP  and  OM  P,  PM  =  PM 


'.A  LMP  is  congruent  to  A  OMP 
.'.  LM  =  PO  and  LP  =  MO 


Ex.  1.'  If  two  perpendiculars  are  drawn  to  the     R 
upper  base  of  a  parallelogram  from  the  extremities 
of  the  lower  base,  two  congruent  right  triangles  are 
formed. 


(Iden.). 
(66). 

CO- 
(76). 
(27). 

Q.E.D. 


Ex.  2.   The  angles  adjoining  each  base  of  an 
isosceles  trapezoid  are  equal. 

Ex.  3.   If  the  angles  at  the  base  of  a  trapezoid 
are  equal,  the  figure  is  isosceles. 

Ex.  4.   The  diagonals  of  a  rectangle  are  equal. 

Ex.  6.   If  the  diagonals  of  a  parallelogram  are  equal, 
the  figure  is  a  rectangle. 

Ex.  6.    Any  line  terminated  in  a  pair  of  oppo-       R 
site  sides  of  a  parallelogram  and  passing  through 
the  midpoint  of  a  diagonal  is  bisected  by  this 
point. 

A  D 

Ex.   7.   If  one  angle  of  a  parallelogram  is  a  right  angle,  the  figure  is  a 
rectangle. 

Ex.   8.    The  bisectors  of  the  angles  of  a  trapezoid  form  a  quadrilateral, 
two  of  whose  angles  are  right  angles. 


QUADRILATERALS 


49 


Ex.  9.  The  bisectors  of  the  four  interior 
angles  formed  by  a  transversal  cutting  two 
parallels  form  a  rectangle. 

[Prove  each  Z  of  LMPQ  a  rt.  Z.] 


M 


Ex.  10.   The  bisectors  of  the  angles  of  a  parallelogram  form  a  rec- 
tangle. 

Ex.  11.  The  bisectors  of  the  angles  of  a  rectangle 
form  a  square.  [In  order  to  prove  EFGH  equilat- 
eral, the  /&  A  HB  and  CDF  are  proved  congruent 
and  isosceles ;  similarly  &BGC  and  AED.~\ 


Ex.  12.  The  perpendiculars  upon  a  diagonal  of  a 
parallelogram  from  the  opposite  vertices  are  equal. 


Ex.  13.  If  on  diagonal  BD,  of  square  A  BCD, 
BE  is  taken  equal  to  a  side  of  the  square,  and 
EP  is  drawn  perpendicular  to  BD  meeting  AD  at  P, 
AP  =  PE  =  ED. 


Ex.  14.  If  AB  CD  is  a  square  and  E,  F,  G,  H  are 
points  on  the  sides,  such  that  AE  =  BF=CG=DH, 
EFGH  is  a  square. 

[First,  prove  EFGH  equilateral ;  then  one  Z  a  rt.  Z.] 


Ex.  16.   The  diagonals  of  an  isosceles  trapezoid  are  equal. 

125.  COROLLARY.    Parallel  lines  included  between  parallel 
lines  are  equal.  (124.) 

126.  COROLLARY.    The  diagonal  of  a  parallelogram  divides 
it  into  two  congruent  triangles. 

127.  COROLLARY.    The  opposite  angles  of  a  parallelogram 
are  equal.  (27.) 


50  BOOK   I.     PLANE   GEOMETRY 

PROPOSITION  XXXIX.     THEOREM 

128.  If  the  opposite  sides  of  a  quadrilateral  are  equal,  the 
figure  is  a  parallelogram.     [Converse  of  124.] 

Given:   Quadrilateral  ABCD ; 
AB  =  DC',    AD  =  BC. 

To  Prove :  ABCD  is  a  O. 

Proof  :  Draw  diagonal  BD.  £ 

In  A  ABD  and  CBD,  BD  =  BD 

AB  ==  DC  and  AD  =  BC 
.'.A  ABD  is  congruent  to  A  CBD 

.'.  AB  is  II  to  DC 
Also  Z  y  =  Z  x 

.'.  AD  is  II  to  BC 
Hence  ABCD  is  a  O 

PROPOSITION  XL.     THEOREM 

129.  If  two  sides  of  a  quadrilateral  are  equal  and  parallel, 
the  figure  is  a  parallelogram. 

Given:  Quadrilateral  ABCD  ;  AB  =  DC  and  AB  II  to  DC. 
To  Prove  :  ABCD  is  a  O. 
Proof:  Draw  diagonal  BD. 

In  A  ABD  and  CBD,        BD  =  BD  (Iden.). 

AB  =  DC  (Given). 

Za  =  Zi  (66). 

.-.A  ABD  is  congruent  to  A  CBD  (52). 

Hence  Z.y=.£x  (27). 

.'.  AD  is  II  to  BC  (70). 

.'.  ABCD  is  a  O  (Def.).   Q.E.D. 

130.  COROLLARY.     Any  pah"  of  consecutive  angles  of  a  paral- 
lelogram are  supplementary.  (69.) 


QUADRILATERALS  51 

PROPOSITION  XL  I.     THEOREM 

131.    The  diagonals  of  a  parallelo-         p 
gram  bisect  each  other. 

Given :  O  EFGH ;  diagonals  EG  and 
FH  intersecting  at  X. 


To  Prove :  FX  =  XH  and  GX  =  XE.       E  H 

Proof :  In  A  FXG  and  EXH,         FG  =  EH  (124). 

Za  =  Zo  and  ^c  =  /_r  (66). 

.'.  A  FXG  is  congruent  to  A  EXH  (76). 

.'.  FX  =  XII  and  GX  =  XE         (27).  Q.E.D. 

PROPOSITION  XLII.     THEOREM 

132.  If  the  diagonals  of  a  quadrilateral  bisect  each  other, 
the  figure  is  a  parallelogram. 

Given:  Quadrilateral  EFGH,  diagonals  EG  and  FH, 
FX  =  XH,  EX  =  GX. 

To  Prove :  EFGH  is  a  O. 

Proof :  In  A  FXG  and  EXH,  FX  =  XH  and  EX  =  XG  (Hyp. ). 

Z  FXG  =  Z  EXH  (?). 

.*.  A  FXG  is  congruent  to  A  EXH  (?). 

.-.  FG  =  EH  and  Z  <?  =  Z  r  (?). 

/.  ra  is  II  to  .E#  (70)- 

.'.EFGH  is  fill]  (129).    Q.E.D. 


Ex.  1.   The  lines  joining  a  pair  of  opposite  vertices  of  a  parallelogram 

to  the  midpoints  of  the  opposite  sides  are  equal  ^ F  E 

and  parallel. 

Ex.  2.  If  through  the  point  of  intersection 
of  the  diagonals  of  a  parallelogram,  two  lines 
are  drawn  intersecting  a  pair  of  opposite  sides  B  C" 

(produced  if  necessary),  the  intercepts  on  these  sides  are  equal. 

Ex.  3.  It  is  impossible  to  draw  two  straight  lines  from  the  ends  of 
the  base  of  a  triangle  terminating  in  the  opposite  sides,  so  that  they  shall 
bisect  each  other. 


52  BOOK   I.     PLANE   GEOMETRY 

PROPOSITION  XLIII.     THEOREM 

133.  Two  parallelograms  are  congruent  if  two  sides  and  the 
included  angle  of  one  are  equal  respectively  to  two  sides  and 
the  included  angle  of  the  other. 


A  D  O 

Given  :  UJ  AC  and  LN  ;  AB  =  LM  ;  AD  =  LO  ;  Z.A  =  Z.  L. 
To  Prove  :  The  UJ  are  congruent. 

Proof:  Superpose  O  AC  upon  O  LN,  so  that  the  equal 
A  A  and  L  coincide,  AD  falling  along  LO  and  AB  along  LM. 
Point  D  coincides  with  point  O  [AD  =  LO  (Hyp.)]. 

Point  B  coincides  with  point  M  [AB  =  LM  (Hyp.)]. 

BC  and  MN  are  both  II  to  LO  (Def.). 

.*.  BC  falls  along  MN  (Ax.  13). 

CD  and  NO  are  both  II  to  LM  (Def.). 

.'.CD  falls  along  NO  (?). 

Hence  C  falls  exactly  upon  N  (38). 

.*.  the  figures  coincide,  and  are  congruent     (26).     Q.E.D. 

134.  COROLLARY.    Two  rectangles  are  congruent  if  the  base 
and  the  altitude  of  one  are  equal  respectively  to  the  base  and 
the  altitude  of  the  other. 

PROPOSITION  XLIV.     THEOREM 

135.  The  diagonals  of  a  rhombus  (or  of  a  square)  are  per- 
pendicular to  each  other,  bisect  each  other,  and  bisect  the 

angles  of  the  rhombus  (or  of  the  square). 

p 

Given:  Rhombus  ABCD-,  diagonals  AC,  BD. 

To  Prove:  AC  _L  to  BD-,  AC  and  BD 
bisect  each  other  ;  and  they  bisect  A  DAB, 
ABC,  etc. 


QUADRILATERALS  53 

Proof :  Point  A  is  equally  distant  from  B  and  D  (122). 

Point  C  is  equally  distant  from  B  and  D  (?). 

.-.^Cis±to^D  (83).  Q.E.D. 

Also  AC  and  BD  bisect  each  other     (131).  Q.E.D. 

Also  DB  bisects  Z  ADC  and  Z  ABC  (85). 

Similarly,      AC  bisects  Z  D4£  and  Z  DCS  (85). 

Q.E.D. 

PROPOSITION  XLV.     THEOREM 

136.  The  line  joining  the  midpoints  of  two  sides  of  a  tri- 
angle is  parallel  to  the  third  side  B 
and  equal  to  half  of  it.                                                Sk 

Given:   A  ABC-,  M,  the  midpoint    p  M>f 

of  AB  ;  P,  the  midpoint  of  BC ;  line     \ 
MP. 
To   Prove:    MP  II  to  AC;  A 

and   MP  =  ^  ^1C. 

Proof :     Suppose  ^1R  is  drawn  through  4,  II  to  BC  and 
meeting  3fP  produced,  at  E. 

In  A  ABM  and  BPJtf,      -4lf  =  BM  (Hyp.)- 

Z^  =  Ze  (51). 

Z  o  =  Z.B  (66). 

.'.  A  ^.Rjtf  is  congruent  to  A  BPM  (76). 

.'.AR  =  BP  (27). 

But                                  BP  =  PC  (Hyp.)- 

.'.AR  =  PC  (Ax.  1). 

.-.^ICPBisaO  (129). 

Hence                   UP,  or  JfP,  is  II  to  AC  (Def.).  Q.E.D. 

Also                                   RP=AC  (124). 

But                                  MP  =  BM  (27). 

/.  MP,  the  half  of  HP,  =  %  AC  (Ax.  6). 

Q.E.D. 


54 


BOOK   I.   'PLANE   GEOMETRY 


PROPOSITION  XLVI.     THEOREM 

137.  The  line  bisecting  one  side  of  a 
triangle  and  parallel  to  a  second  side, 
bisects  also  the  third  side. 

Given:  A  ABC;  MP  bisecting  AB  and 
II  to  AC. 

To  Prove :  MP  bisects  BC  also. 
Proof:  Suppose  MX  is  drawn  from  Jf,  the  midpoint  of  AB 
to  X,  the  midpoint  of  BC. 

MX  is  II  to  AC  (136). 

But  MP  is  II  to  AC  (Hyp.). 

.*.  MX  and  MP  coincide  (Ax.  13). 

That  is,  MP  bisects  BC.  Q.E.D. 


Ex.  1.  The  lines  joining  the  midpoints  of  the 
sides  of  a  triangle  divide  the  triangle  into  four 
congruent  triangles. 


Ex.  2.     The  lines  joining  (in  order)  the  midpoints  of  the  sides 
quadrilateral   form   a   parallelogram   the  sum   of 
whose  sides  is  equal  to  the  sum  of  the  diagonals  of 
the  quadrilateral. 

Ex.  3.  The  lines  joining  (in  order)  the  mid- 
points of  the  sides  of  a  rectangle  form  a  rhombus. 
[Draw  the  diagonals.] 

Ex.  4.  If  the  four  midpoints  of  the  four 
halves  of  the  diagonals  of  a  parallelogram 
are  joined  in  order,  another  parallelogram  is 
formed. 

Prove  this  in  four  ways. 

Ex.  6.  If  lines  are  drawn  from  a  pair  of 
opposite  vertices  of  a  parallelogram  to  the 
midpoints  of  a  pair  of  opposite  sides,  they 
trisect  the  diagonal  joining  the  other  two 
vertices. 


QUADRILATERALS 


55 


Ex.  6.  If  two  medians  are  drawn  from  two 
vertices  of  a  triangle  and  produced  their  own 
length  beyond  the  opposite  sides,  and  if  these  ex- 
tremities are  joined  to  the  third  vertex,  these 
two  lines  are  equal,  and  in  the  same  straight 
line. 


Ex.  7.  The  line  joining  the  midpoints  of  one 
pair  of  opposite  sides  of  a  quadrilateral  and  the 
line  joining  the  midpoints  of  the  diagonals 
bisect  each  other. 


PROPOSITION  XL VII.     THEOREM 

138.  The  line  bisecting  one  leg  of  a  trapezoid  and  parallel 
to  the  base  bisects  the  other  leg,  is  the  median,  and  is  equal 
to  half  the  sum  of  the  bases. 

Given:  Trapezoid  ABCD;  Jtf,  the  midpoint  of  AB;  MP  II  to 
AD,  meeting  CD  at  P.  B  C 

To  Prove :  /\T 

I.    P  is  the  midpoint  of  CD.  M  /  R 

II.    MP  is  the  median.  Y 

III.     MP  =  1(AD  +  BC).  *£- aD 

Proof:     I.    Draw  diagonal  BD,  meeting  MP  at  R. 

MP  is  II  to  BC  (63). 

InA^iBD,  MR  bisects  BD  (137). 

In  A  BbC,  RP  bisects  CD  (?). 

That  is,  P  is  the  midpoint  of  CD.  Q.E.D. 

II.    MP  is  the  median  (Def.  123).     Q.E.D. 

III.    IllA^BD,  MR  =  I  AD  (136). 

Also,  in  A  BDC,  RP  =  ^  BC (?). 

Adding,  .-.  MP=  \(AD+  BC)  (Ax.  2). 

Q.E.D. 


56 


BOOK   I.     PLANE   GEOMETRY 


139.   COROLLARY.     The  median  of  a  trapezoid  is  parallel  to 
the  bases  and  equal  to  half  their  sum. 

Ex.  1.  In  a  trapezoid  one  of  whose  bases  is 
double  the  other,  the  diagonals  intersect  at  a 
point  two  thirds  of  the  distance  from  each  end 
of  the  longer  base  to  the  opposite  vertex. 

Proof :  Take  M,  the  midpoint  of  A  0  (136),  etc. 

Ex.  2.  If  one  angle  of  a  triangle  is  double  an- 
other, the  line  from  the  third  vertex,  making  with 
the  longer  adjacent  side  an  angle  equal  to  the  less 
given  angle,  divides  the  triangle  into  two  isosceles  triangles. 

NOTE.  The  verb  "  to  intersect "  means  merely 
"  to  cut."  In  geometry,  the  verb  "  to  intercept " 
means  "  to  include  between"  Thus  the  statement 
"  AB  and  CD  intercept  XY  on  the  line  EF"  really 
means,  "AB  and  CD  intersect  EF&nd  include  XY, 
a  part  of  EF,  between  them." 


/ 


f 


PROPOSITION  XL VIII.    THEOREM 

140.  Parallels  intercepting  equal 
parts  on  one  transversal  intercept 
equal  parts  on  any  transversal. 

Given :  I  Is  AB,  CD,  EF,  GH,  U  inter- 
cepting equal  parts  AC,  CE,  EG,  GI,  on 
the  transversal  AI,  and  cutting  trans- 
versal BJ. 

To  Prove  :   BD  =  DF  =  FH  =  HJ. 

Proof:    The  figure  ABFE  is  a  trapezoid 

CD  bisects  AE  and  is  ||  to  EF 
.-.  D  is  midpoint  of  BF 

That  is,  BD  =  DF. 

Similarly,  CDHG  is  a  trapezoid  and  DF=  FH. 

Similarly,  FH=HJ.      . ' .  BD  =  DF=  FH=  HJ 


(?), 

(Hyp.). 
(?). 


(Ax.  1). 

Q.E.D. 


QUADRILATERALS 


57 


PROPOSITION  XLIX.     THEOREM 

141.  The  midpoint  of  the  hypotenuse  of  a 
right  triangle  is  equally  distant  from  the  three 
vertices. 

Given :  Rt.  A  ABC ;  Jf,  the  midpoint  of  the 
hypotenuse  AB. 

To  Prove :   AM  =  CM  =  BM. 


Proof :    Suppose  MX  drawn  n  to  BC,  meeting  AC  at  X. 
X  is  the  midpoint  of  AC 

MX  is  .L  to  AC  (64). 

.'.  AM  =  CM  (80). 

But  AM=BM  (Hyp.). 

.*.  AM  =  CM  =  BM  (Ax.  1). 

Q.E.D. 

Ex.  1.  Any  right  triangle  can  be  divided  by  one  line  into  two  isosceles 
triangles. 

Ex.  2.   If  through  the  vertex  of  the  right  p Q      i 

angle  of  a  right  triangle  a  line  is  drawn 
parallel  to  the  hypotenuse,  the  legs  of  the 
right  triangle  bisect  the  angles  formed  by 
this  parallel  and  the  median  drawn  to  the 
hypotenuse. 

Ex.  3.  If  one  leg  of  a  trapezoid  is  perpendicular  to  the  bases,  the  mid- 
point of  the  other  leg  is  equally  distant  from  the  ends  of  the  first  leg. 
[Draw  the  median.] 

Ex.  4.   The  median  of  a  trapezoid  bisects  both  the  diagonals. 

Ex.  5.   The  line  joining  the  midpoints  of  the  diagonals  of  a  trapezoid 
is  a  part  of  the  median,  is  parallel  to  the  bases, 
and  is  equal  to  half  their  difference. 

Ex.  6.  If  the  median  of  a  triangle  is  equal  to 
half  the  side  to  which  it  is  drawn,  it  is  a  right 
triangle. 

Ex.  7.  The  line  (prolonged  if  necessary)  joining  the  midpoints  of  two 
sides  of  a  triangle,  bisects  the  altitude  drawn  to  the  third  side. 

ROBBINS'S    NEW    PLANE    GEOM.  —  6 


M 


58  BOOK   I.     PLANE   GEOMETRY 

Ex.  8.   If  one  acute  angle  of  a  right  triangle  is  double  the  other,  the 
hypotenuse  is  double  the  shorter  leg. 

Proof:  Use  fig.  of  141.     Denote  £A  by  x, 

then                                                 Z  B  =  2  x  and  x  =  30°  (?). 

AM=MC  =  BM  (141). 

.-.  Z  BCM  =  Z  B  =  60°  (55). 

.-.  Z  £MC  =  60°  (104). 

.-.  MB  =  BC  (115). 

.-.  AB,  or  2  x  M£,  =  2  x  C£  (Ax.  6). 


PROPOSITION  L.     THEOREM 

142.   The  perpendiculars  from  the  vertices  of  a  triangle  to 
the  opposite  sides  meet  in  a  point. 

R,-  C 


A   .     Z 


Given:  A  ABC,  AX  J_  to  BC,  £F-L  to  ^ic,  and  cz_L  to  AB. 
To  Prove :  These  three  J§  meet  in  a  point. 
Proof :  Through  A  suppose  BS  drawn  II  to  BC ;  through  B, 
TS  II  to  AC;  through  C,  ET  II  to  AB,  forming  AEST. 

The  figure  ABCR  is  a  O  (Const). 

and  ABTC  is  a  O  (?). 

BC  =  AB  and  CT  =  AB  (124) . 

.'.BC=CT  (Ax.  1). 

Now  CZ  is  _L  to  #r  (64). 

That  is,  CZ  is  _L  bisector  of  BT. 

Similarly,  AX  is  J_  bisector  of  E-S. 

And  BY  is  _L  bisector  of  TS. 

.-.  in  A  BST,  AX,  BY,  CZ  meet  at  a  point        (100). 

Q.E.D. 


QUADRILATERALS 


59 


PROPOSITION  LI.     THEOREM 

143.  The  three  medians  of  a  triangle  meet  in  a  point  which 
is  two  thirds  the  distance  from  any  vertex  to  the  midpoint  of 
the  opposite  side. 

Given :  A  ABC,  medians  AF,  BD  and 
CE,  the  latter  two  meeting  at  O. 
(Fig.  1.) 

To  Prove :  BO  =  f  BD  ;   co  =  |  CE  ; 

AO  =  |  AF  and  that  all  three  medians    B F  c 

meet  at  o.  FlG<  L 

Proof :  Suppose  H  is  midpoint  of  BO  and  z  is  the  midpoint 
of  CO.  Draw  ED,  DI,  IH,  HE. 

In  A  ABC,  ED  is  II  to  BC  and  =  \  BC  (136). 

In  A  OBC,  HI  is  II  to  BC  and  =  ± BC  (136). 

.-.  ED  =  HI  (Ax.  1);  and  ED  is  II  to  HI  (63). 

.-.  EDIH  is  a  O  (129). 

.-.  HO  =  OD  and  IO  =  OE  (131). 

.-.  BH  =  HO  =  OD  and   CI  =  IO  =  OE        (Ax.  1). 

That  is,  BO  =  f  BD  and  CO  =  f  CE. 

Suppose  AF  meets  BD  at  Or .     (Fig.  2.) 

Then     BO=%BD  (Proved  above). 

And     BOr=%BD         (Proved  similarly). 

.-.  BO  =BOf  (Ax.  1). 

That  is,  O  and  o'  are  the  same  point,  and 

the  three  medians  meet  at  O,  which  is  J  the 

distance  from  any  vertex  to  the  midpoint  of  the  opposite 

side.  Q.E.D. 


F  • 
FIG.  2. 


Ex.  1.  If  two  lines  (AB  and  CD)  are  equal 
and  parallel,  the  lines  connecting  their  op- 
posite ends  bisect  each  other. 

Ex.  2.  If  the  bisector  of  one  angle  of  a  tri- 
angle is  perpendicular  to  the  opposite  side,  the 
triangle  is  isosceles. 


B 


60  BOOK   I.     PLANE   GEOMETRY 


Ex.  3.   The  median  of  an  isosceles  triangle  is  perpen- 
dicular to  the  base. 

Ex.  4.    The  medians  from  the  ends  of  the  base  of  an 
isosceles  triangle  are  equal. 


Ex.  6.   If  a  triangle  has  two  equal  medians,  it  is 
isosceles. 


A  B 

Ex.  6.  If  ABC  is  an  equilateral  triangle  and  D,  E,  F  are  points  on 
the  sides,  such  that  AD  =  BE  =  CF,  triangle  DEF  is  also  equilateral. 

B 

Ex.  7.   Any  two   vertices  of  a  triangle  are 
equally  distant  from  the  median  from  the  third 

vertex. 

f\  ••"  \ 

Y 

Ex.  8.  The  lines  bisecting  two  interior  angles  that  a  transversal 
makes  with  one  of  two  parallels  cut  off  equal  segments  on  the  other 
parallel  from  the  point  at  which  the  transversal  meets  it.  [The  & 
formed  are  isosceles.] 

POLYGONS 

144.  A   polygon  is   a  portion    of   a   plane    bounded    by 
straight  lines.     The  lines  are  called  the  sides.     The  points 
of  intersection  of  the  sides  are  the  vertices.     The  angles  of 
a  polygon  are  the  angles  at  the  vertices. 

145.  The  number  of  sides  of  a  polygon  is  the  same  as  the 
number  of   its  vertices  or  the  number  of   its   angles.     An 
exterior  angle  of  a  polygon  is  an  angle  without  the  polygon, 
between  one  side  of  the  polygon  and  another  side  prolonged. 

146.  An  equilateral  polygon  has  all  its  sides  equal  to  one 
another.     An  equiangular  polygon  has  all  its  angles  equal  to 
one  another. 


POLYGONS  61 

147.  A  convex  polygon  is  a  polygon  no  side  of  which  if 
produced  will  enter  the  surface  bounded  by  the  sides  of  the 
polygon.  A  concave  polygon  is  a  polygon  at  least  two  sides  of 
which  if  produced  will  enter  the  polygon. 


EQUILATERAL  EQUIANGULAR  CONCAVE,  OR 

CONVEX  POLYGONS  REENTRANT 

NOTE.  A  polygon  may  be  equilateral  and  not  be  equiangular;  or 
it  may  be  equiangular  and  not  be  equilateral.  The  word  "polygon" 
usually  signifies  convex  figures. 

148.  Two  polygons  are  mutually  equiangular  if  for  every 
angle  of  the  one  there  is  an  equal  angle  in  the  other  and 
similarly   placed.     Two   polygons   are   mutually  equilateral, 
if  for  every  side  of  the  one  there  is  an  equal  side  in  the 
other,  and  similarly  placed. 

149.  Homologous   angles   in    two    mutually   equiangular 
polygons  are  the  pairs  of  equal  angles.     Homologous  sides 
in  two  polygons  are  the  sides  between  two  pairs  of  homolo- 
gous angles. 

150.  Two  polygons  are  congruent  if  they  are  mutually  equi- 
angular and  their  homologous  sides  are  equal ;  or  if  they  are 
composed   of   triangles,    equal   each  to   each   and  similarly 
placed.     (Because  in  either  case  the  polygons  can  be  made 
to  coincide.)  

Ex.  1.  If  a  quadrilateral  has  three  equal  sides,  is  it  necessarily  a 
parallelogram  ?  a  trapezoid  ? 

Ex.  2.  Two  quadrilaterals  are  congruent  if  three  sides  and  the  two 
included  angles  of  one  are  equal  respectively  to  three  corresponding  sides 
and  the  two  included  angles  of  the  other. 


62 


BOOK   I.     PLANE   GEOMETRY 


161.  Two  polygons  may  be  mutually  equiangular  without 
being  mutually  equilateral  ;  also,  they  may  be  mutually 
equilateral  without  being  mutually  equiangular  —  except  in 
the  case  of  triangles. 


The  first  two  figures  are  mutually  equilateral,  but  not  mutually  equi- 
angular. The  last  two  figures  are  mutually  equiangular,  but  not  mutu- 
ally equilateral. 

152.  A  3-sided  polygon  is  a  triangle. 

A  4-sided  polygon  is  a  quadrilateral. 
A  5-sided  polygon  is  a  pentagon. 
A  6-sided  polygon  is  a  hexagon. 
A  7-sided  polygon  is  a  heptagon. 
An  8-sided  polygon  is  an  octagon. 
A  10-sided  polygon  is  a  decagon. 
A  12-sided  polygon  is  a  dodecagon. 
A  15-sided  polygon  is  a  pentadecagon. 
An  w-sided  polygon  is  called  an  n-gon. 

PROPOSITION  LII.    THEOREM 

153.  The  sum  of  the  interior  angles 
of  an  n-gon  is  equal  to  (n  —  2)  times 
180°. 

Given :  A  polygon  having  n  sides. 

To  Prove :   The  sum  of   its  interior 
A=  (n-2).  180°. 

Proof :  If  all  possible  diagonals  are  drawn  from  any  vertex 
it  is  evident  that  there  are  formed  (n  —  2)  triangles. 

The  sum  of  the  A  of  one  A  =  180°  (104). 


ANGLES  63 

/.  the  sum  of  the  A  of  (n  -  2)  A  =  (n  -  2)  180°  (Ax.  3). 
The  sum  of  A  of  the  A  =  sum  of  A  of  n-gon  (Ax.  4). 
/.  the  sum  of  A  of  the  n-gon  —  (n  —  2)  •  180°  (Ax.  1).  Q.E.D. 

154.  COROLLARY.     The  sum  of  the  interior  angles  of  an  n- 
gon  is  equal  to  180°  •  n  —  360°. 

155.  COROLLARY.     Each    angle    of    an   equiangular    n-gon 
=  Qi  -  2)  180° 

n 

156.  COROLLARY.    The  sum  of  the  angles  of  any  quadrilat- 
eral is  equal  to  four  right  angles. 

157.  COROLLARY.     If   three  angles  of  a  quadrilateral  are 
right  angles,  the  figure  is  a  rectangle. 

PROPOSITION  LIII.     THEOREM 

158.  If  the  sides  of  a  polygon  are  produced  hi  order,  one  at 
each  vertex,  the  sum  of  the  exterior  angles  of  the  polygon  equals 
four  right  angles,  that 

is,  360°. 

Given:  A  polygon 
with  sides  prolonged 
in  succession  forming 
the  several  exterior 
angles  &,  6,  <?,  c?,  etc. 

To  Prove  :Za  +  Z£  +  Z<?  +  Zd  +  etc.  =  4  rt.  A  =  360°. 

Proof :  Suppose  at  any  point  in  the  plane,  lines  are  drawn 
parallel  to  the  several  sides  of  the  given  polygon,  extending 
in  the  same  direction,  and  forming  A  A,  B,  C,  Z>,  etc. 

Then  Z^t  +  Zs  +  Zc+ZD  +  Z^-f  etc.  =  4 rt.  A    (47). 

But  Z^t  =  Za,  Z£  =  Z£,  Zc  =  Z<?,  ZD  =  ZC?,  etc.     (74). 

Substituting, 

etc.  =  4  rt.  A  =  360°     (Ax.  6). 

Q.E.D. 


64  BOOK   I.     PLANE   GEOMETRY 

159.    COROLLARY.     Each  exterior  angle   of  an  equiangular 
polygon  is  equal  to  —  —  s,  that  is,  to 


160.    COROLLARY.     The  sum  of  the  exterior  angles  of  a  poly- 
gon is  independent  of  the  number  of  its  sides. 


Ex.  1.  How  many  degrees  are  there  in  the  sum  of  all  the  angles  of  a 
pentagon  ?  of  a  decagon  ?  of  a  dodecagon  ? 

Ex.  2.  How  many  degrees  are  there  in  each  angle  of  an  equiangular 
hexagon  ?  of  an  equiangular  octagon  ? 

Ex.  3.  How  many  degrees  are  there  in  each  exterior  angle  of  an 
equiangular  pentagon  ?  of  an  equiangular  hexagon  ?  16-gon  ? 

Ex.  4.  How  many  sides  has  the  polygon  the  sum  of  whose  interior 
angles  exceeds  the  sum  of  its  exterior  angles  by  900°  ? 

Ex.  6.    The  sum  of  the  angles  at  the  vertices  of  a  . 

five-pointed  star  (pentagram)  is  equal  to  two  right  A 

angles.  /  \ 

Ex.  6.  If  two  angles  of  a  quadrilateral  are  supple- 
mentary, the  other  two  are  supplementary. 

Ex.  7.  If  from  any  point  within  an  angle  per- 
pendiculars to  the  sides  are  drawn,  they  include 
an  angle  which  is  the  supplement  of  the  given  angle. 

Ex.  8.   If  a  quadrilateral  has  four  equal  sides,  what  kind  is  it? 
Ex.  9.   Can  a  trapezoid  have  three  equal  sides?  a  trapezium? 

Ex.  10.  A  colt  is  tied,  by  a  chain  30  ft.  long,  to  the  four  corners  of  a 
lot  60  ft.  square,  on  four  successive  days.  Using  a  scale  of  £  in.  to  the 
foot,  draw  a  diagram  showing  the  area  over  which  the  colt  grazes  during 
these  four  days.  Draw  another  diagram,  using  the  same  scale,  for  a 
chain  15  ft.  long;  40  ft.  long. 

Ex.  11.  A  calf  is  tied  by  a  chain  20  ft.  long,  to  the  four  corners  of  a 
barn  40  ft.  square,  on  four  successive  days  and  allowed  to  graze  in  the 
adjoining  pasture.  Using  a  scale  of  \  in.  to  the  foot,  draw  a  diagram 
showing  the  area  grazed  over  at  the  end  of  the  fourth  day.  Draw  another 
diagram,  using  the  same  scale,  for  a  chain  15  ft.  long ;  25  ft.  long. 


SYMMETRY 


65 


SYMMETRY 

161.  A  figure  is  symmetrical  with  respect  to  a  line  if,  by 
using  that  line  as  an  axis,  the  part  of  the  figure  on  one  side 
of  the  line  may  be  folded  over,  and  will   exactly  coincide 
with  the  part   on   the  other  side.     This  line  is  an  axis  of 
symmetry. 

162.  A  figure  is  symmetrical  with   respect  to  a  point  if 
this  point  bisects  every  line  drawn  through  it  and  terminated 
(both  ways)  in  the  boundary  of  the  figure. 

This  point  is  the  center  of  symmetry. 

163.  It  is  evident  that  the  axis  of  symmetry  bisects  at 
right  angles  every  line  joining  two  symmetrical  points  ;  and 
that  the  center  of  symmetry  bisects  every  line  joining  any 
pair  of  points  symmetrical  with  respect  to  it. 


Examples  of  symmetry  are  given  in  these  figures. 

First  figure  is  symmetrical  with  respect  to  XX'  as  an  axis.  (Why?) 

Second  figure  is  symmetrical  with  respect  to  0  as  a  center.  (Why  ?) 

P  and  P/  are  symmetrical  with  respect  to  XX'  as  an  axis.  (Why?) 
A  and  A ',  B  and  B',  etc.  are  symmetrical  with  respect  to  0  as  a  center. 

(Why?)       XX'  is  _L  to  P'P[  and  bisects  it.      AO  =  A'O,  BO  =B'0, 

etc.  

Ex.  1.  Is  the  altitude  of  an  isosceles  triangle  an  axis  of  symmetry  ? 
the  altitude  of  a  scalene  triangle? 

Ex.  2.   Is  the  diagonal  of    a  rhomboid  an  axis  of  symmetry?  of  a 
square? 

Ex.3.   Has  any  triangle  a  center  of  symmetry?  a  parallelogram  ? 


66  BOOK   I.     PLANE   GEOMETRY 

PROPOSITION  LIV.     THEOREM 

164.  If  two  lines  are  symmetrical  with  respect  to  a  center, 
they  are  equal  and  parallel. 

Given :  AB  and  RS  symmetrical  with 
respect  to  o,  that  is,  every  line  through 
O,  terminated  in  AB  and  RS,  is  bisected 
at  o ;  A8  and  BR,  two  such  lines. 
To  Prove :  AB  =  RS  and  AB  II  to  RS. 

Proof :  Draw  AR  and  BS.     Ao  =  OS  and  BO  =  OR  (Hyp.). 

.-.  ABSR  is  a  O  (132). 

.-.  AB  =  RS  (124). 

Also  AB  is  II  to  RS  (Def.  120). 

Q.E.D. 

PROPOSITION  LV.    THEOREM 

165.  If  a  diagonal  of  a  quadrilateral 
bisects  two  of  its  angles,  this  diag- 
onal is  an  axis  of  symmetry. 

Given :  Quadrilateral  ABCD  ;  AC  a 
diagonal  bisecting  /.BAD  and  Z  BCD. 
To  Prove :  ABCD  symmetrical  with 
respect  to  AC. 

Proof :  In  A  ABC  and  ADC,  AC  —  AC  (?). 

Z  BAG  =  Z  DAG  and  Z  BCA  =  Z  DCA  (?). 

.  •.  A  ABC  is  congruent  to  A  ADC  (?). 

.•.  AC  is  an  axis  of  symmetry  (161). 

Q.E.D. 

166.  COROLLARY.    The  diagonal  of  a  square  or  of  a  rhombus 
is  an  axis  of  symmetry. 


Ex.   The  point  of  intersection  of  the  diagonals  of  a  parallelogram  is  a 
center  of  symmetry.     Prove. 


SYMMETRY 


67 


N 


PROPOSITION  LVI.    THEOREM 

167.  If  a  figure  is  symmetrical  with  respect  to  two  perpen- 
dicular axes,  it  is  symmetrical  with 
respect  to  their  intersection  as  a  center. 

Given:  Figure  MN  symmetrical 
with  respect  to  the  J_  axes  XX1  and 
YYf  which  intersect  at  o. 

To  Prove  :  Figure  MN  is  symmetri- 
cal with  respect  to  O  as  a  center. 

Proof  :  Take  any  point  P  in  the  boundary.  Draw  PB  _L 
to  rr',  intersecting  TY1  at  A  and  meeting  the  boundary  at 
B.  Draw  BE  _L  to  XX1  ,  intersecting  XX1  at  G  and  meet- 
ing the  boundary  at  E.  Draw  AC,  OP,  OE. 

[The  demonstration  is  accomplished  by  proving  POE  a 
straight  line,  bisected  at  O.] 

PB  is  II  to  XX!  and  BE  is  II  to  YYr 
.-.  ABCO  is  a,  CJ 

.'.BC=AO 
But  BC=CE 

.'.  AO  —  CE  (Ax.  1). 

Hence                           ACEO  is  a  O  (129). 

.-.  EO=CA  (124). 

Also                             EO  is  II  to  CA  (120). 
Similarly,          ACOP  may  be  proved  a  O. 

.•.  PO  is  =  to  AC  and  II  to  AC. 

Hence                    POE  is  a  straight  line  (Ax.  13). 

And                                   PO  =  EO  (Ax.  1). 
But  P  is  any  point  in  the  boundary,  so  POE  is  any  line 
through  O. 

.*.  O  is  a  center  of  symmetry  (161). 

Q.E.D. 


(62). 

(Def.). 

(124). 


Ex.   Prove  that  no  triangle  can  have  a  center  of  symmetry. 


68  BOOK   I.    PLANE   GEOMETRY 

CONCERNING  ORIGINAL   EXERCISES 

168.  In  the  original  work  which  this  text  contains,  the 
pupil  is  expected  to  state  the  hypothesis  and  the  conclusion 
of  each  theorem,  and  to  apply  them  to  an  appropriate  figure ; 
also  to  give  a  complete  and  logical  statement  of  the  proof, 
with  a  reason  for  every  statement. 

In  many  of  these  exercises,  suggestions  are  made  and  such 
assistance  is  given  as  experience  has  shown  to  be  needed  by 
average  pupils.  This  is  done  in  order  to  encourage  definite 
accomplishment,  which  is  one  of  the  greatest  incentives  to 
further  effort. 

To  apply  the  knowledge  acquired  from  the  preceding  pages 
is  now  the  student's  task.  His  interest  in  this  science  will 
depend  largely  on  the  success  of  his  efforts  to  prove  originals. 

The  student  should  not  draw  a  special  figure  for  a  general 
proposition.  That  is,  if  "  triangle  "  is  specified,  he  should 
draw  a  scalene  and  not  an  isosceles  or  a  right  triangle  ;  and 
if  "  quadrilateral "  is  mentioned,  he  should  draw  a  trapezium 
and  not  a  parallelogram  or  a  square. 

SUMMARY.     GENERAL  DIRECTIONS  FOR  ATTACKING 
EXERCISES 

169.  A  triangle  is  proved  isosceles  by  showing  that  it  con- 
tains two  equal  sides,  or  two  equal  angles. 

170.  A  triangle  is  proved  a  right  triangle  by  showing  that 
one  of  its  angles  is  a  right  angle,  or  that  two  of  its  angles  are 
complementary,  or  that  one  of  its  angles  is  equal  to  the  sum 
of  the  other  two. 

171.  Right  triangles   are  proved   congruent   by  showing 
that  they  have  : 

(1)  Hypotenuse  and  acute  angle  of  one  equal  etc. 

(2)  Hypotenuse  and  leg  of  one  equal  etc. 

(3)  The  legs  of  one  equal  etc. 

(4)  Leg  and  adjoining  angle  of  one  equal  etc. 

(5)  Leg  and  opposite  angle  of  one  equal  etc. 


ORIGINAL  EXERCISES  69 

172.  Oblique  triangles  are  proved  congruent  by  showing 
that  they  have  : 

(1)  Two  sides  and  the  included  angle  of  one  equal  etc. 

(2)  One  side  and  the  adjoining  angles  of  one  equal  etc. 

(3)  Three  sides  of  one  equal  etc. 

173.  Angles  are  proved  equal  by  showing  that  they  are  : 

(1)  Equal  to  the  same  or  to  equal  angles. 

(2)  Halves  or  doubles  of  equals. 

(3)  Vertical  angles. 

(4)  Complements  or  supplements  of  equals. 

(5)  Homologous  parts  of  congruent  figures. 

(6)  Base  angles  of  an  isosceles  triangle. 

(7)  Corresponding  angles,  alternate  interior  angles,  etc.,  of  parallels. 

(8)  Angles  whose  sides  are  respectively  parallel  or  perpendicular. 

(9)  Third  angles  of  triangles  which  have  two  angles  of  one  equal  etc. 

174.  Lines  are  proved  equal  by  showing  that  they  are  : 

(1)  Equal  to  the  same  or  to  equal  lines. 

(2)  Halves  or  doubles  of  equals. 

(3)  Distances  to  the  ends  of  a  line  from  any  point  in  its  perpendicu- 

lar bisector. 

(4)  Homologous  parts  of  congruent  figures. 

(5)  Sides  of  an  isosceles  triangle. 

(6)  Distances  to  the  sides  of  an  angle  from  any  point  in  its  bisector. 

(7)  Opposite  sides  of  a  parallelogram. 

(8)  The  parts  of  one  diagonal  of  a  parallelogram  made  by  the  other. 

175.  Two  lines  are  proved  perpendicular  by  showing  that 
they: 

(1)  Make  equal  adjacent  angles  with  each  other. 

(2)  Are  legs  of  a  right  triangle. 

(3)  Have  two  points  in  one,  each  equally  distant  from  the  ends  of  the 

other. 

176.  Two  lines  are  proved  parallel  by: 

(1)  The  customary  angle  relations  of  parallel  lines. 

(2)  Showing  that  they  are  opposite  sides  of  a  parallelogram. 

(3)  Showing  that  they  are  parallel  or  perpendicular  to  a  third  line. 


TO  BOOK   I.     PLANE   GEOMETRY 

177.    Two  lines,  or  two  angles,  are  proved  unequal  by  the 
usual  axioms  and  theorems  pertaining  to  inequalities. 

[See  especially,  Ax.  5;  Ax.  12;  81,  86,  87,  88,  III,  91,  92,  103,  116, 
118.] 

ORIGINAL  EXERCISES 

A  f* 

1.   Parallel  lines  are  everywhere  equally  distant.         - 
Given:  \\s  AC  and  BD;  AB  and  CD  Js  to  AC. 
To  Prove :  AB  =  CD. 


2.  If  two  lines  in  a  plane  are  everywhere  equally  distant,  they  are 
parallel. 

3.  The  bisectors  of  any  two  consecutive  angles  of  a  parallelogram 
meet  at  right  angles. 

4.  The  line  drawn  from  any  point  in  the  base  of  an  isosceles  triangle 
to  the  opposite  vertex  is  less  than  either  leg. 

6.  If  ABC  is  an  equilateral  triangle  and  each  side  is  produced  (in 
order)  the  same  distance,  so  that  AD  =  BE  =  CF,  the  triangle  DEF 
is  equilateral. 

6.  If  A  BCD  is  a  square  and  the  sides  are  produced  (in  order)  the  same 
distance,  so  that  AE  =  BF  =  CG  =  DH,  the  figure  EFGH  is  a  square. 

7.  The  two  lines  joining  the  midpoints  of  the  opposite  sides  of  a 
quadrilateral  bisect  each  other.     [Join  the  4  midpoints  (in  order),  etc.] 

8.  If  two  adjacent  angles  of  a  quadrilateral  are  right  angles,  the 
bisectors  of  the  other  angles  are  perpendicular  to  each  other. 

9.  If  two  opposite  angles  of  a  quadrilateral  are  right  angles,  the 
bisectors  of  the  other  angles  are  parallel. 

10.   Two  isosceles  triangles  are  congruent,  if: 

(1)  The  base  and  one  of  the  adjoining  angles  in  the  one  are  equal 
respectively  to  the  base  and  one  of  the  adjoining  angles  in  the  other. 

(2)  A  leg  and  one  of  the  base  angles  in  the  one  are  equal  respectively 
to  a  leg  and  one  of  the  base  angles  in  the  other. 

(3)  The  base  and  vertex  angle  in  one  are  equal  to  the  same  in  the 
other. 

(4)  A  leg  and  vertex  angle  in  one  are  equal  to  the  same  in  the  other. 

(5)  A  leg  and  the  base  in  one  are  equal  to  the  same  in  the  other. 


ORIGINAL   EXERCISES  71 


11.  If  upon  tha  three  sides  of  any  triangle  equi- 
lateral triangles  are  constructed  (externally)  and  a 
line  is  drawn  from  each  vertex  of  the  given  tri- 
angle to  the  farthest  vertex  of  the  opposite  equi- 
lateral triangle,  these  three  lines  are  equal. 

Proof  :  Z  EA  C  =  Z  BA  F  (?).  Add  to  each  of 
these  Z  CAB. 


Then  prove  A  EAB  and  CAF  congruent. 
Similarly,  A  CAD  is  congruent  to  A  CEB.    Etc. 

12.  The  sum  of  the  diagonals  of  any  quadrilateral  is  less  than  the 
sum  of  the  four  sides,  but  greater  than  half  that  sum. 

13.  The  difference  between  two  sides  of  a  triangle  is  less  than  the 
third  side. 

14.  The  bisectors  of  the  exterior  angles  of  a  rectangle  form  a  square. 
16.    The  bisectors  of  the  equal  angles  of  an  isosceles  triangle  (termi- 
nating in  the  equal  sides)  are  equal. 

16.  The  median  to  the  base  of  an  isosceles  triangle  bisects  the  vertex 
angle. 

17.  The  perpendiculars  to  the  legs  of  an  isosceles  triangle  from  the  mid- 
point of  the  base  are  equal. 

18.  State  and  prove  the  converse  of  Ex.  17. 

19.  If  AB  =  LM  and  AL  =  BM,  ZB  =  ZL 
and  /.  EA  0  =  /.  OML  and  BO  =  OL. 

20.  The  bisectors  of  a  pair  of  corresponding     A*"~  """ M 
angles  are  parallel. 

21.  If  two  lines  are  cut  by  a  transversal  and  the  exterior  angles  on 
the    same    side    of    the  transversal   are   supplementary,   the   lines    are 
parallel. 

22.  The  bisectors  of  a  pair  of  vertical  angles  are  in  the  same  straight 
line. 

23.  The  midpoint  of  a  diagonal  of  a  parallelo- 
gram is  a  center  of  symmetry. 

24.  If  the  base  angles  of  a  triangle  are  bisected 
and  through  the  intersection  of  the  bisectors  a  line  is 
drawn  parallel  to  the  base  and  terminating  in  the 
sides,  this  line  is  equal  to  the  sum  of  the  parts  of 
the  sides  it  meets,  between  it  and  the  base. 


72  BOOK  I.     PLANE   GEOMETRY 

25.  In    two    congruent    triangles,  homologous   medians   are   equal; 
homologous  altitudes  are  equal ;  homologous  bisectors  are  equal. 

26.  If  two  parallel  lines  are  cut  by  a  transversal,  the  two  exterior 
angles  on  the  same  side  of  the  transversal  are  supplementary. 

27.  If  from  a  point  a  perpendicular  is  drawn 
to  each  of  two  parallels,  they  are  in  the  same  line. 
[Draw  a  third  II  through  the  point.] 

28.  The  median  to  one  side  of  a  triangle  is 

less  than  half  the  sum  of  the  other  two  sides.  """-....  \      / 

Proof :  Produce  median  EM  to  R,  so  that  MR  "VR 

=  BM,  draw  AR  and  CR.     Fig.isO.    (?).    Etc. 

29.  The  sum  of  the  medians  of  a  triangle  is  less  than  the  sum  of  the 
sides  of  the  triangle. 

30.  If  the  diagonals  of  a  trapezoid  are 
equal,  it  is  isosceles. 

[Draw  DR   and  CS  J_  to  AB;  and 
prove  rt.  &  A  CS  and  BDR  congruent,  to       . 

__  A  p^  ^  *—f 

31.  If  a  perpendicular  is  drawn  from  each  vertex  of  a  parallelogram  to 
any  line  outside  the  parallelogram,  the  sum  of  the  Js  from  one  pair  of 
opposite  vertices  equals  the  sum  of  the  Js  from  the  other  pair. 

32.  The  sum  of  the  perpendiculars  to  the  legs  of 
an    isosceles  triangle  from  any  point  in  the   base 
equals  the  altitude  upon  one  of  the  legs.     (That  is, 
the  sum  of   the  perpendiculars  from   any  point  in 
the  base  of  an  isosceles  triangle  to  the  equal  sides  re- 
mains a  uniform  length  for  every  point  of  the  base.) 

[Prove  PE  =  CF.'] 

33.  The  sum  of  the  three  perpendiculars  drawn    A 

from  any  point  within  an  equilateral  triangle,  to  the  three  sides,  re- 
mains a  uniform  length  for  all  positions  of  the  point. 

[Draw  a  line  through  this  point  ||  to  one  side ;  draw  the  altitude  of 
the  A  J_  to  this  line  and  side ;  prove  the  sum  of  the  three  Js  equals  this  alti- 
tude and  hence  equals  a  constant.] 

34.  If  from  any  point  in  the  base  of  an  isosceles  tri- 
angle parallels  to  the  equal  sides  are  drawn,  the  sum 
of  the  sides  of  the  parallelogram  formed  is  equal  to 
the  sum  of  the  legs  of  the  triangle. 


ORIGINAL   EXERCISES  73 

35.    The    bisector  of    the  right   angle    of    a 

right  triangle  is  also  the  bisector  of  the  angle 

formed  by  the  median  and  the  altitude  drawn 

from  the  same  vertex. 

To  Prove  :  Z  MCS  =  Z  LCS.  A  M  S  L 

Proof:  ZACS  =  ^BCS  (?)  ;  ZACM  =  ZBCL  (?).    Now  use  Ax.  2. 

36.  If  the  vertex  angle  of  an  isosceles  triangle  is  equal  to  the  sum 
of  the  base  angles,  any  line  perpendicular  to  the  base  forms  with  the 
sides  of   the  given  triangle  (one   side  to   be  produced)   three  isosceles 
right  triangles. 

37.  If  two  sides  of  a  triangle  are  unequal  and 
the  median  to  the  third  side  is  drawn,  the  angles 
formed  with  the  base  are  unequal. 


38.    State  and  prove  the  converse  of  Ex.  37. 


.A  M  C 

39.  If  the  opposite  sides  of  a  hexagon  are  equal  and  parallel,  the 
three  diagonals  drawn  between  opposite  vertices  meet  in  a  point. 

40.  In  triangle  ABC,  AD  is  perpendicular  to  BC,  meeting  it  at  D ; 
E  is  the  midpoint  of  AB,  and  F  of  AC',  the  angle  EDF  is  equal  to  the 
angle  EAF. 

41.  If  the  diagonals  of  a  quadrilateral  are  equal,  and  also  one  pair 
of  opposite  sides,  two  of  the  four  triangles  into  which  the  quadrilateral 
is  divided  by  the  diagonals  are  isosceles. 

42.  If  angle  A  of  triangle  ABC  equals  three  times  angle  B,  there  can 
be  drawn  a  line  AD  meeting  BC  in  D,  such  that  the  triangles  ABD  and 
A  CD  are  isosceles. 

43.  If  E  is  the  midpoint  of  side  BC  of  parallelogram  ABCD,  AE 
and  BD  meet  at  a  point  two  thirds  the  distance  from  A  to  E  and  from 
D  to  B. 

44.  If  in  triangle  ABC,  in  which  AB  is  not  equal  to  A  C,  AC'  is  taken 
on  AB  (produced  if  necessary)  equal  to  AC,  and  AB'  is  taken  on  AC 
(produced  if  necessary)  equal  to  AB,  and  B'C'  is  drawn  meeting  BC 
at  D,  then  AD  bisects  angle  BAG. 

Proof:  A  ABC  is  congruent  to  &AB'C'  (?)  (52).  .-.their  homolo- 
gous parts  are  equal.  Thus  prove  that  A  BC'D  is  congruent  to  kB'CD. 
Etc. 

45.  If  a  diagonal  of  a  parallelogram  bisects  one  angle,  it  also  bisects 
the  opposite  angle. 

ROBBINS'S    NEW    PLANE    GEOM. — 6 


74  BOOK    I.     PLANE   GEOMETRY 

47.  If  a  diagonal  of  a  parallelogram  bisects  one  angle,  the  figure  is 
equilateral. 

48.  Any  line  drawn  through  the  point  of  intersection  of  the  diagonals 
of  a  parallelogram  divides  the  figure  into  two  congruent  quadrilaterals. 

49.  If  AR  bisects  angle  A  of  triangle  ABC  and  AT  bisects  the  ex- 
terior angle  at  A,  any  line  parallel  to  AB,  having  its  extremities  in  AR 
and  A  T,  is  bisected  by  A  C. 

60.  If  the  opposite  angles  of  a  quadrilateral  are  equal,  the  figure  is  a 
parallelogram. 

51.  If,  in  isosceles  triangle  XYZ,  AD  is  drawn  from  A,  the  midpoint 
of   YZ,  perpendicular  to  the  base  XZ,  DZ  =  J  XZ.     [Draw  alt.  from 
Y.-] 

52.  If  ABC  is  an  equilateral  triangle,  if  the  bisectors  of  angles  B 
and  C  meet  at  D,  if  DE  is  drawn  parallel  to  AB  meeting  AC  at  E,  and 
DF,  parallel  to  BC  meeting  A  C  at  F,  then  A  E  =  ED  =  EF  =  DF  =  CF. 

63.  If  A  is  any  point  in  RS  of  triangle  RS T,  and  B  is  the  midpoint 
of  RA,  C  the  midpoint  of  AS,  D  the  midpoint  of  ST,  and  E  the  mid- 
point of  TR,  then  BCDE  is  a  parallelogram. 

54.  If  lines  are  drawn  from  any  vertex  of  a  parallelogram  to  the  mid- 
points of  the  two  opposite  sides,  they  divide  the  diagonal  which  they 
intersect  into  three  equal  parts. 

Proof:  Draw  the  other  diagonal. 

55.  If  the  interior  and  exterior  angles  at  two  vertices  of  a  triangle  are 
bisected,  a  quadrilateral  is  formed,  having  two  of  its  angles  right  angles 
and  the  other  two  supplementary. 

56.  The  four  bisectors  of  the  angles  of  a  quadrilateral  form  a  second 
quadrilateral  whose  opposite  angles  are  supplementary. 

Proof:  Extend  a  pair  of  opposite  sides  of  the  given  quadrilateral  to 
meet  at  X.  Bisect  the  base  angles  of  the  new  A  formed,  meeting  at  0. 
Then  show  that  Z  0  equals  one  of  the  A  between  the  given  bisectors,  and 
Z  0  is  supplementary  to  the  angle  opposite. 


BOOK  II 
THE   CIRCLE 

178.  A  curved  line  is  a  line  no  part  of  which  is  straight. 

179.  A  circle  is  a  plane  curve  all  points  of   which   are 
equally  distant  from  a  point  in  the  plane,  called  the  center. 

180.  The  length  of  the  circle  is  called  the  circumference. 

181.  A  radius  is  a  straight  line  drawn  from  the  center  to 
any  point  of  the  circle. 

A  diameter  is  a  straight  line  that  contains  the  center,  and 
the  extremities  of  which  are  in  the  circle. 


CIRCLE  SECANT  CHORD  CENTRAL    ANGLE  SEMI- 

RADIUS  TANGENT  INSCRIBED   ANGLE      CIRCUMFERENCES 

DIAMETER  POINT   OF    CONTACT  ARC  SEMICIRCLES 

A  secant  is  a  straight  line  cutting  the  circle  in  two  points. 

A  chord  is  a  straight  line  the  extremities  of  which  are 
in  the  circle. 

A  tangent  is  a  straight  line  which  touches  the  circle  at 
only  one  point,  and  does  not  cut  it,  however  far  it  may  be 
extended.  The  point  at  which  the  line  touches  the  circle  is 
called  the  point  of  contact  or  the  point  of  tangency. 

A  common  tangent  to  two  circles  is  a  line  tangent  to  both 
of  them. 

75 


76  BOOK   II.     PLANE   GEOMETRY 

182.  A  central  angle  is  an  angle  formed  by  two  radii. 

An  inscribed  angle  is  an  angle  whose  vertex  is  on   the 
circle  and  whose  sides  are  chords. 

183.  An  arc  is  any  part  of  a  circle. 

A  semicircle  is  an  arc  equal  to  half  a  circle. 
A  quadrant  is  an  arc  equal  to  one  fourth  of  a  circle. 
Equal  circles  are  circles  having  equal  radii. 
Concentric  circles  are  circles  having  the  same  center. 


SECTOR  CIRCLES    INTERNALLY        CIRCLES   EXTERNALLY 

SEGMENT  TANGENT  TANGENT 

NOTE.  A  circle  is  named  either  by  its  center  or  by  three  of  its  points 
as  "the  O  0"  or  "the  O  ABC." 

184.  A  sector  is  the  figure   bounded  by  two  radii   and 
their  included  arc. 

A  segment  is  the  figure  bounded  by  an  arc  and  its  chord. 

185.  Two  circles  are  tangent  to  each  other  if  they  are  tan- 
gent to  the  same  line  at  the  same  point.     Circles  may  be 
tangent  to  each  other  internally,  if   the  one  is  within   the 
other,  or  externally,  if  each  is  without  the  other. 

186.  POSTULATE.    A  circle  can  be  described  about  any  given 
point  as  center  and  with  any  given  line  as  radius. 

Subtend  is  used  in  the  sense  of  "to  cut  off."  A  chord 
subtends  an  arc.  Hence  an  arc  is  subtended  by  a  chord. 

An  angle  is  said  to  intercept  the  arc  between  its  sides. 
Hence  an  arc  is  intercepted  by  an  angle. 

The  hypothesis  is  contained  in  what  constitutes  the  sub- 
ject of  the  principal  verb  of  the  theorem. 


THE   CIRCLE 


187. 


PRELIMINARY  THEOREMS 
THEOREM.    All  radii  of   the   same   circle  are   equal. 


(179.) 

188.  THEOREM.    All  radii  of  equal  circles  are  equal.    (183.) 

189.  THEOREM.    The  diameter  of  a  circle  equals  twice  the 
radius. 

190.  THEOREM.    All  diameters  of  the  same  or  of  equal  cir- 
cles are  equal.  (Ax.  3.) 

191.  THEOREM.    The  diameter  of  a  circle  bisects  the  circle. 
Given  :  Any  O  and  a  diameter. 

To  Prove  :  The  diameter  bisects  the  circle. 

Proof :  Suppose  one  segment  folded  over  upon  the  other 

segment,  using  the  diameter  as  an  axis.     If  the  arcs  do  not 

coincide,  there  are  points  of   the    circle  unequally  distant 

from  the  center.     But  this  is  impossible  (179). 

.*.  the  arcs  coincide  and  are  equal  (26). 

Q.E.D. 

192.  THEOREM.   With  a  given  point  as  center  and  a  given 
line  as  radius,  it  is  possible  to  describe  only  one  circle.  (179.) 

That  is,  a  circle  is  determined  if  its  center  and  radius  are 
fixed. 

Historical  Note.  Archimedes  was  born  at  Syracuse,  Sicily,  during  the 
third  century  B.C.  He  is  regarded  as  the  greatest 
mathematician  of  antiquity,  and  probably  of  all 
time,  save  only  that  modern  wizard,  Sir  Isaac 
Newton.  He  was  educated  in  Egypt  and  won  the 
respect  and  admiration  of  the  king,  Hiero,  for  his 
exceptionaLgenius  in  the  construction  of  mechan- 
ical devices  and  mathematical  formulas.  These 
included  the  measurement  of  the  circle  (circum- 
ference and  area),  the  cone,  the  cylinder,  and  the 
sphere.  To  him  is  given  credit  also  for  the  dis- 
covery of  specific  gravity.  Cicero  relates  his  discovery  of  the  tomb  of 
Archimedes  over  a  century  after  his  burial  in  Syracuse. 


ARCHIMEDES 


T8  BOOK   II.     PLANE   GEOMETRY 

THEOREMS  AND  DEMONSTRATIONS 
PROPOSITION  I.     THEOREM 

193.  In  the  same  circle  (or  in  equal  circles)  equal  central 
angles  intercept  equal  arcs. 


Given :  O  o  =  O  C  ;  Z  o  =  Z  C. 
To  Prove :  Arc  AB  =  arc  LM. 

Proof :    Superpose  0  O  upon  the  equal  O  C,  making   Z  o 

coincide  with  its  equal,  Z  C.     Point  A  falls  on  i,  and  point 

B  on  M  (188). 

Arc  AB  coincides  with  arc  LM  (179). 

.'.  arc  AB  =  arc  LM  (26). 

Q.E.D. 

PROPOSITION  II.     THEOREM 

194.  In  the  same  circle  (or  in  equal  circles)  equal  arcs  are 
intercepted  by  equal  central  angles.  [Converse.] 

Given :  O  O  =  0  C  ;  arc  AB  =  arc  LM. 
To  Prove :  Z  o  =  Z  C. 

Proof :  Superpose  O  O  upon  the  equal  O  C,  making  the 
centers  coincide.  Point  A  falls  on  point  L.  Then  arc  AB 
coincides  with  arc  LM  and  point  B  falls  on  point  M.  (Because 
the  arcs  are  equal.) 

.•.  OA  coincides  with  Ci,  and  OB  with  CM        (39). 
/.  ZO  =  ZC  (26). 

Q.E.D. 


THE   CIKCLE  79 

PROPOSITION  III.     THEOREM 
195.   In  the  same  circle  (or  in  equal  circles)  : 

I.  If  two  central  angles  are  unequal,  the  greater  angle  inter- 
cepts the  greater  arc. 

II.  If  two  arcs  are  unequal,  the  greater  arc  is  intercepted 
by  the  greater  central  angle.     [Converse.] 


I.  Given :  O  o  =  O  C ;  Z  LCM  >  Z  o. 
To  Prove :  Arc  LM  >  arc  AB. 

Proof :  Superpose  O  O  upon  O  C,  making  sector  AOB  fall 
in  position  of  sector  XCM,  OB  coinciding  with  CM. 

CX  is  within  the  angle  LCM.     (Because  Z  LCM  >  Z  o.) 

Arc  AB  falls  upon  LM,  in  the  position  XM       (179). 

.'.  arc  LM  >  arc  XM  (Ax.  5). 

That  is,  arc  LM  >  arc  AB.  Q.E.D. 

II.  Given:  (?). 

To  Prove :  Z  LCM  >  ^  o. 

Proof :  The  pupil  may  employ  either  superposition,  as  in  I, 
or  the  method  of  exclusion,  as  in  92. 

NOTE.  Unless  otherwise  specified,  the  arc  of  a  chord  always  refers  to 
the  lesser  of  the  two  arcs.  If  two  arcs  (in  the  same  or  equal  circles)  are 
concerned,  it  is  understood  either  that  each  is  less  than  a  semicircle,  or 
each  is  greater. 


Ex.  1.   Are  equal  circles  also  congruent?    Why? 

Ex.  2.   Is  there  a  geometrical  figure  that  is  both  sector  and  segment? 


80  BOOK   II.     PLANE   GEOMETRY 

PROPOSITION  IV.     THEOREM 

196.   In  the  same  circle  (or  in  equal  circles)  equal  chords 
subtend  equal  arcs. 


Given :  O  o  =  O  C ;  chord  AB  =  chord  LM. 
To  Prove :  Arc  AB  =  arc  LM. 

Proof :   Draw  the  several  radii  to  the  ends  of  the  chords. 

In  A  OAB  and  CLM,  OA  =  CL  and  OB  =  CM  (188). 

Chord  AB  =  chord  LM  (Hyp.). 

.-.  A  OAB  is  congruent  to  A  CLM  (?). 

Hence                               Zo  =  Zc  (?). 

.-.  arc  AB  =  arc  LM  (193). 

Q.E.D. 

PROPOSITION  V.     THEOREM 

197.  In  the  same  circle  (or  hi  equal  circles)  equal  arcs  are 
subtended  by  equal  chords.     [Converse.] 
Given :   O  o  =  O  C  ;  arc  AB  =  arc  LM. 
To  Prove:  Chord  AB  =  chord  LM. 

Proof :   Draw  the  several  radii  to  the  ends  of  the  chords. 

In  A  OAB  and  CLM,  OA  =    CL  (188). 

OB=  CM  (80). 

Zo  =  ZC  (194). 

.  •.  A  AOB  is  congruent  to  A  CLM  (?). 

.-.  chord  AB  =  chord  LM  (?). 

Q.E.D. 


THE   CIRCLE  81 

PROPOSITION  VI.     THEOREM 

198.    In  the  same  circle  (or  in  equal  circles)  : 

I.  If  two  chords  are  unequal,  the  greater  chord  subtends  the 
greater  arc. 

II.  If  two  arcs  are  unequal,  the  greater  arc  is  subtended  by 
the  greater  chord.     [Converse.] 


I.  Given :  O  0  =  O  C ;  chord  AB  >  chord  RS. 
To  Prove :  Arc  AB  >  arc  RS. 

Proof :  Draw  the  several  radii  to  the  ends  of  the  chords. 
In  A  AOB  and  RCS, 

AO  =  EC  and  BO  =  SC  (?). 

Chord  AB  >  chord  B8  (Hyp.). 

.-.  Z  O  >  Z  C  (92). 

.•.  arc  AB  >  arc  RS  (195,  I). 

Q.E.D. 

II.  Given :  O  o  =  O  C ;  arc  AB  >  arc  RS. 
To  Prove :  Chord  AB  >  chord  RS. 
Proof :  Draw  the  several  radii. 

In  A  AOB  and  RCS, 

AO  =  EC  and  BO  =  SC  (?). 

But  Z  o  >  Z  c  (195,  II). 

.-.  chord  AB  >  chord  RS  (91). 

Q.E.D. 

Ex.   Can  either  part  of  Proposition  VI  be  proved  by  the  method  of  ex- 
clusion ?    Can  Proposition  IV  or  V  be  proved  by  that  method  ? 


82 


BOOK   II.     PLANE   GEOMETRY 


PROPOSITION  VII.     THEOREM 

199.  The  diameter  perpendicular  to  a 
chord  bisects  the  chord  and  both  the  sub- 
tended arcs. 

Given :  Diameter  DR  _L  to  chord  AB  in 
00. 

To  Prove : 
I.  AM  =  MB  ; 

II.  Arc  AB  =  arc  RB,  arc  AD  =  arc  DB. 
Proof :  Draw  radii  to  ends  of  the  chord. 
I.  In  rt.  A  OAM  and  OBM, 

OA  =  OB 
OM=  OM 

A  OAM  is  congruent  to  A  OBM 
.'.  AM=  BM 


II. 

Also 


Z  AOM  =  Z  BOM 

arc  AR  =  arc  RB 
Z  AOD  =  Z  BOD 
arc  -4D  =  arc  DB 


(84). 
CO- 

Q.E.D. 

(27). 
(193). 

(49). 
(193). 

Q.E.D. 


200.  COROLLARY.     The    line  from   the  center  of  a  circle 
perpendicular  to  a  chord  bisects  the  chord. 

201.  COROLLARY.     The   perpendicular  bisector  of   a  chord 
passes  through  the  center  of  the  circle. 

Proof  :  The  center  is  equally  distant  from  the  extremities 

of  the  chord  (187). 

.•.  the  center  is  in  the  JL  bisector  of  the  chord  (82). 


Ex.  1.   The  perpendicular  bisectors   of   all  chords  in   a 
circle  pass  through  a  common  point. 


THE   CIRCLE 


88 


Ex.  2.   A  diameter  bisecting  a  chord  is  perpendicular  to 
the  chord  and  bisects  the  subtended  arcs. 


Ex.  3.    A  diameter  bisecting  an  arc  is  the  perpendicular  bisector  of 
the  chord  of  the  arc. 


Ex.  4.   A  line  bisecting  a  chord  and  its  arc  is  the     B 
perpendicular  bisector  of  the  chord. 


Ex.  6.   If  a  circle  is  described  on  the  hypotenuse  of  a  right  triangle 
as  diameter,  it  passes  through  the  vertex  of  the  right  angle  (141). 

Ex.  6.   If  any  number  of  parallel  chords  are  drawn  in  a  circle,  their 
midpoints  all  lie  on  the  same  straight  line. 

Ex.  7.    If  two  perpendicular  diameters  of  a  circle  are  drawn  and  their 
extremities  are  joined  in  order,  these  chords  form  a  square. 

Ex.  8.   If  any  two  diameters  of  a  circle  are  drawn  and  their  extremities 
are  joined  in  order,  the  figure  is  a  parallelogram. 


PROPOSITION  VIII.     THEOREM 

202.  The  line  perpendicular  to  a 
radius  at  its  extremity  is  tangent  to 
the  circle. 

Given:  Radius  OA  of  O  O,  and 
ET  JL  to  OA  at  A. 

To  Prove:  ET  tangent  to  the 
circle. 

Proof :  Take  any  point  P  in  ET  (except  A)  and  draw  OP. 

OP  >  OA  (87). 

.  •.  P  lies  without  the  O  (Because  OP  >  radius). 

That  is,  eve'ry  point  (except  A)  in  ET  is  without  the  O. 

.•.  ET  is  a  tangent  to  the  O  O  (Def.). 

Q.E.D. 


84 


BOOK   II.     PLANE   GEOMETRY 


PROPOSITION  IX.     THEOREM 

203.  If  a  line  is  tangent  to  a  circle,  the  radius  drawn  to  the 
point  of  contact  is  perpendicular  to  the  tangent.     [Converse.] 

Given:  ET  tangent  to  O  O  at  A  ; 
radius  OA. 

To  Prove :  OA  J_  to  ET. 

Proof :  Every  point  (except  A)  in 
ET  is  without  the  O  (181). 

.•.a  line  from  O  to  any  point  in 
ET  (except^)  is  >  OA.  (Because  it 
is  >  a  radius.) 

That  is,  OA  is  the  shortest  line  from  O  to  ET. 

.-.  OA  is_L  to  ET  (87). 

204.  COROLLARY.  The  perpendicular  to  a  tangent  at  the  point 
of  contact  passes  through  the  center  of  the  circle.  (43.) 

Q.E.D. 

PROPOSITION  X.     THEOREM 

205.  If  two  circles  are  tangent 
to  each  other,  the  line  joining 
their    centers    passes    through 
then*  point  of  contact. 

Given :  ©  O  and  C  tangent  to 
a  line  at  A,  and  line  OC. 

To  Prove :  OC  passes  through  A. 
Proof:  Draw  radii  OA  and  CA. 


(203). 
(48). 
(39). 

Q.E.D. 
Let  the  pupil  supply  the  proof  if  the  circles  are  tangent  internally. 


OA  is  _L  to  the  tangent  and  CA  is  J_  to  the  tangent 

.-.  OAC  is  a  straight  line 
.*.  OAC  and  OC  coincide,  and  OC  passes  through  A 


THE   CIRCLE 


85 


PROPOSITION  XL     THEOREM 

206.  Two  tangents  drawn  to  a  circle  from  an  external  point 
are  equal. 


NOTE.     In  this  theorem  the  word  "tangent"  signifies  the  distance 
between  the  external  point  and  the  point  of  contact. 

Given :  O  O  and  tangents  PA,  PB. 

To  Prove :  Distance  PA  =  distance  PB. 

Proof :  Draw  radii  to  points  of  contact,  and  join  OP. 

Z  OAP  and  OBP  are  right  A  (203). 

In  rt.  A  OAP  and  OBP,  OP  =  OP  (?)  ;   OA  =  OB  (?). 

.*.  A  O^iP  is  congruent  to  A  OBP  (84). 

.'.PA  =  PB  (?). 

Q.E.D. 

Historical  Note.  Pythagoras,  a  Greek  philosopher,  born  probably  at 
Samos,  in  the  sixth  century  B.C.,  had  the  reputation  of  being  an  "  assiduous 
inquirer,"  and  of  having  a  great  fund  of  general 
knowledge.  He  was  a  moral  reformer  as  well  as  a 
scientific  teacher.  He  was  the  head  of  a  secret 
society  the  members  of  which  were  pledged  to  the 
severest  discipline,  and  to  the  practice  of  temper- 
ance, purity,  and  obedience.  The  study  of  mathe- 
matics in  Greece  was  magnified  by  him  and  ad- 
vanced to  the  rank  of  a  science.  The  invincible 
proof  given  on  page  204  of  the  theorem  that  the 
square  on  the  hypotenuse  of  a  right  triangle  is  equal 

in  area  to  the  sum  of  the  squares  on  the  legs,  has  been  attributed  to 
Pythagoras,  and  is  often  referred  to  as  the  Pythagorean  proposition. 
The  discovery  of  incommensurable  magnitudes  (p.  96)  and  of  many  other 
theorems  and  problems  is  ascribed  to  him. 


PYTHAGORAS 


86  BOOK  II.     PLANE   GEOMETRY 

PROPOSITION  XII..    THEOREM 

207.  If  from  an  external  point  tangents  are  drawn  to  a  circle, 
and  radii  are  drawn  to  the  points  of  contact,  the  line  joining  the 
center  and  the  external  point  bisects : 
I.  The  angle  formed  by  the  tangents. 

n.  The  angle  formed  by  the  radii. 

HI.  The  chord  joining  the  points  of  contact. 

IV.  The  arc  intercepted  by  the  tangents. 


Given:  Tangents  AP  and  BP  from  point  P  and  radii  OA 
and  OB. 

To  Prove  :  Line  OP  bisects  : 

I.  Z  APB,  II.   Z  AOB,  III.    Chord  AB,  IV.   Arc  AXE. 
Proof  :  A  OAP  and  OBP  are  rt.  A  (?). 

They  are  congruent.     (Explain.) 


II. 

III.  O  is  equidistant  from  A  and  B  (?). 
P  is  also  equidistant  from  A  and  B  (206). 

/.  OP  is  -L  to  AB  at  its  midpoint  (83). 

IV.  Arc  AX  =  arc  BX  (193). 

Q.E.D. 

Ex.  1.  Tangents  drawn  to  a  circle  at  the  extremities  of  a  diameter 
are  parallel. 

Ex.  2.  Tangents  drawn  to  a  circle  at  the  extremities  of  a  chord  form, 
with  the  chord,  an  isosceles  triangle. 

Ex.  3.  The  bisector  of  the  angle  between  two  tangents  to  a  circle 
passes  through  the  center. 


THE   CIRCLE 


Ex.  4.  The  sum  of  one  pair  of  opposite  sides  of  a 
circumscribed  quadrilateral  is  equal  to  the  sum  of  the 
other  pair.  A 

Ex.  5.   A  circumscribed  parallelogram  is  equilateral. 
Ex.  6.   A  circumscribed  rectangle  is  a  square. 


Ex.  7.  If  a  circle  is  inscribed  in  a  right  triangle, 
the  sum  of  the  diameter  and  the  hypotenuse  is  equal 
to  the  sum  of  the  legs. 

Ex.  8.  If  two  parallel  tangents  meet  a  third  tangent, 
and  lines  are  drawn  from  the  points  of  intersection  to 
the  center,  they  are  perpendicular. 

Ex.  9.  Tangents  drawn  to  two  tangent  circles  from 
any  point  in  their  common  interior  tangent  are  equal. 

Ex.  10.  The  common  interior  tangent  of  two  tan- 
gent circles  bisects  their  common  exterior  tangent. 

Ex.  11.  Do  the  theorems  of  Ex.  9  and  10  apply  if 
the  circles  are  tangent  internally?  If  so,  prove. 

Ex.  12.  In  the  adjoining  figure,  if  AE  and  AD  are 
secants,  AE  passing  through  the  center,  and  the  ex- 
ternal part  of  AD  being  equal  to  a  radius,  the  angle 
DCE  =  Z  /.A. 

[Draw  BC.  /.  DEC  =  ext.  Z  of  &ABC  =  2^A  . 
=  £  D.  (Explain.)  Z  DCE  =  an  ext.  Z,  etc.] 

Ex.  13.    The  two  common  interior  tangents  of  two  circles  are  equal. 


Ex.  14.  The  common 
exterior  tangents  to  two 
circles  are  equal. 

[Produce  them  to  in- 
%  tersection.] 


Ex.  15.   In  the  preceding  figure,  prove  that  RH  —  SF. 
Proof:  AR  +  RB  =  CS  +  SD ; 


XD' 


88 


BOOK   II.     PLANE   GEOMETKY 


...  AR  +  (RH  +  HF}  =  (SF  +  HF)  +  SD. 

.-.  RH  +  RH+  HF=  SF  +  HF  +  SF;  .-.  2  RH  =  2  SF,  etc.  Give 
reasons  and  explain. 

Ex.  16.  The  common  exterior  tangents  to  two  circles  intercept  on  a 
common  interior  tangent  (produced),  a  line  equal  to  a  common  exterior 
tangent.  To  Prove :  RS  =  AB. 

Ex.  17.  AB  and  AC  are  two  tangents  from  A  ;  in  the  less  arc  BC  a 
point  D  is  taken  and  a  tangent  drawn  at  D,  meeting  AB  at  J£  and  AC 
at  F.  Prove  that  AE  +  EF  +  AF  remains  a  uniform  length  for  all  posi- 
tions of  D  in  arc  BC. 

Ex.  18.   If    perpendiculars    are    drawn    upon    a 
tangent  from  the  ends  of  any  diameter : 

(1)  The    point    of    tangency    bisects    the    line 
between  the  feet  of  the  perpendiculars. 

[Draw  CP.] 

(2)  The  sum  of  the  perpendiculars  equals  the 
diameter. 

(3)  The  center  of  the  circle  is  equally  distant  from  the  feet  of  the 
perpendiculars. 


PROPOSITION  XIII.     THEOREM 

208.  In  the  same  circle  (or  in  equal 
circles)  equal  chords  are  equally  distant 
from  the  center. 

Given  :  O  o  ;  chord  AB  =  chord  CD, 
and  distances  OE  and  OF. 

To  Prove  :  OE  =  OF. 

Proof  :  Draw  radii  OA  and  OC. 
In  the  rt.  A  AOE  and  COF, 

AE=  \AB  and  CF= 

But  AB  =  CD 

.'.  AE=  CF 

Also  AO  =  co 

.-.  A  AOE  is  congruent  to  A  COF 
.-.  OE  =  OF 


CD 


(200) 

(HypO 

(Ax.  3) 

(?) 
(84) 

(? 

Q.E.D 


THE   CIRCLE 


89 


PROPOSITION  XIV.     THEOREM 

209.    In  the  same  circle  (or  in  equal  circles)  chords  which 
are  equally  distant  from  the  center  are  equal.     [Converse.] 
Given  :  O  O  ;  chords  AB  and  CD  ;  distance  OE  =  distance  OF. 
To  Prove  :  Chord  AB  =  chord  CD. 
Proof  :  Draw  radii  OA  and  OC. 
In  rt.  A  AOE  and  COF,  AO  =  CO  (?). 

Also                                  EO=OF  (Hyp.). 

.*.  A  AOE  is  congruent  to  A  COF  (84). 

.'.AE=CF  (?). 

Now  AB  is  twice  AE  and  CD  is  twice  CF 


.'.AB=CD 


(200). 
(Ax.  3).    Q.E.D. 


Ex.  1.  If  two  circles  are  concentric,  all  chords  of  the 
greater  that  are  tangent  to  the  less  are  equal. 

Ex.  2.  If  at  the  midpoint  of  an  arc  a  tangent  is  drawn, 
it  is  parallel  to  the  chord  of  the  arc. 

Ex.  3.  If  two  equal  chords  intersect  on  the  circle,  the 
radius  drawn  to  their  point  of  intersection  bisects  their 
angle. 

Ex.  4.  If  the  line  joining  the  point  of  intersection  of 
two  chords  and  the  center  bisects  the  angle  formed  by  the 
chords,  they  are  equal. 

Ex.  5.  The  radius  of  the  circle  inscribed  in  an  equi- 
lateral triangle  is  half  the  radius  of  the  circle  circum- 
scribed about  it.  [Use  143.] 

Ex.  6.  If  the  inscribed  and  circumscribed  circles  of  a 
triangle  are  concentric,  the  triangle  is  equilateral. 

Ex.  7.   If  two  circles  are  concentric  and  a  secant  cuts  them  both,  the 
portions  of  the  secant  intercepted  between  the  circumferences  are  equal. 

Ex.  8.   Of  all  secants  that  can  be  drawn  to  a  circumference  from  a 
fixed  external  point,  the  longest  passes  through  the 
center.  p  ^       OV-"1-^  c- 

Ex.  9.  The  shortest  line  from  an  external 
point  to  a  circumference  is  that  which,  if  produced, 
would  pass  through  the  center. 

ROBBINS'S  NEW  PLANE  GEOM.  —  7 


90 


BOOK   II.     PLANE   GEOMETRY 


PROPOSITION  XV.     THEOREM 

210.  In  the  same  circle  (or  in  equal 
circles)  if  two  chords  are  unequal,  the 
greater  chord  is  at  the  less  distance 
from  the  center. 

Given:  O  O  ;  chord  AB  >  chord  CD, 
and  distances  OE  and  OF. 

To  Prove :  OE  <  OF. 

Proof:  Arc  AB  >  arc  CD  (198,1). 

Suppose  arc  AH  is  taken  on  arc  AB,  equal  to  arc  CD.     Draw 
chord  AH.     Draw  OK  _L  to  J.H,  cutting  AB  at  I. 
Now  chord  AH  =  chord  CD 

.'.  distance  OK  =  distance  OF 
But  OE  <  oi 

Also  01  <  OK 

.*.  OE  <  OK 

Substituting,  OE  <  OF 


(197). 
(208). 
(87). 
(Ax.  5). 
(Ax.  11). 
(Ax.  6). 

Q.E.D. 


PROPOSITION  XVI.     THEOREM 


211.  In  the  same  circle  (or  in  equal  circles)  if  two  chords 
are  unequally  distant  from  the  center,  the  chord  at  the  less 
distance  is  the  greater.  [Converse.] 

Given :  O  O ;  chords  AB  and  CD ;  distance  OE  <  distance  OF. 
To  Prove :  Chord  AB  >  chord  CD. 

Proof:  It  is  evident  that  chord  AB  <  CD,  or  =  chord  CD, 
or  >  chord  CD.  Proceed  by  the  method  of  exclusion. 

Another  Proof :  On  OF  take  OX  =  to  OE.  At  X  draw  a 
chord  RS  J_  to  OX. 

Then                  chord  RS  is  II  to  chord  CD.  (62). 

.-.  arc  RS  >  arc  CD  (Ax.  5). 


THE   CIRCLE 


91 


.-.  chord  ES  >  chord  CD  (198,  II). 
But  chord  AB  =  chord  R8         (209). 

Substituting, 

chord  AB  >  chord  CD    (Ax.  6). 

Q.E.D. 

212.    COROLLARY.     The  diameter  of  a  circle  is  longer  than 
any  other  chord. 

Ex.  1.    What  is  the  longest  chord  that  can  be  drawn  through  a  given 
point  within  a  circle  ? 

Ex.  2.  Of  all  chords  that  can  be  drawn  through  a 
given  point  within  a  circle,  the  chord  perpendicular  to 
the  diameter  through  the  given  point  is  the  shortest. 

Given :  P,  the  point ;  BOC  the  diam. ;  LS  ±  to  EC 
at  P ;  GR,  any  other  chord  through  P. 
To  Prove:  (?) 
Proof:  Draw  OA  ±  to  GR.    Etc. 


PROPOSITION  XVII.     THEOREM 

213.  Through  three  points,  not  in  the 
same  straight  line,  one  circle  can  be 
drawn,  and  only  one. 

Given :   Points  A  and  B  and  c. 
To  Prove:  I.  (?).     II.  (?). 
Proof:   I.    Draw  lines  AB,  BC,  AC.     Suppose  their  _L  bisec- 
tors, OZ,  OX,  OF,  are  drawn.     These  J§  will  meet  at  a  point 

(100). 

With  O  as  a  center  and  OA  or  OB  or  OC  as  a  radius,  a  circle 
can  be  described  through  A,  JB,  and  C  (100). 

II.  These  Ji  meet  at  only  one  point  (100). 

That  is,  there  is  only  one  center. 
The  distances  O^L,  OB,  OC  are  all  equal          (100). 
That  is,  there  is  only  one  radius. 
.*.  there  can  be  only  one  circle          (192).   Q.E.D. 


92 


BOOK   II.     PLANE   GEOMETRY 


214.  COROLLARY.     One  circle,  and  only  one,  can  be  drawn 
through  the  vertices  of  a  triangle. 

215.  COROLLARY.     A  circle  is  determined  by  three  points. 

216.  COROLLARY.     A  circle  cannot  be  drawn  through  three 
points  which  are  in  the  same  straight  line. 

[The  J§  would  be  II.] 

,217.   COROLLARY.     A  straight  line  can  intersect  a  circle  in 
only  two  points.  (216.) 

218.  COROLLARY.     Two   circles   can   intersect  in  only   two 
points. 

PROPOSITION  XVIII. 

219.  If  two  circles  intersect,  the 
line  joining  their  centers  is  the  per- 
pendicular bisector  of  their  common 
chord. 

Given:  (?). 
To  Prove:  (?). 


THEOREM 


Proof 


Draw  radii  in  each  O  to  ends  of  AB. 
Point  O  is  equally  distant  from  A  and  B 
Point  C  is  equally  distant  from  A  and  B 
.-.  OC  is  the  _L  bisector  of  AB 


(1ST). 

CO- 

(83). 

Q.E.D. 


Ex.  1.    Illustrate  the  five  corollaries  on  this  page  by  diagrams. 

Ex.  2.  On  an  island  six  miles  from  the  mainland  is  a  gun  having  a 
range  of  ten  miles.  Draw  a  diagram,  using  a  scale  of  f  in.  to  the  mile, 
showing  the  range  of  the  gun. 

Ex.  3.  On  the  opposite  sides  of  the  entrance  to  a  harbor  are  two  forts, 
twelve  miles  apart.  In  each  there  is  a  gun  with  a  range  of  nine  miles. 
Draw  a  diagram  showing  the  region  exposed  to  the  fire  of  each  gun,  and 
to  the  fire  of  both  guns. 

Ex.  4.  Make  a  similar  problem  using  three  forts,  and  guns  of  different 
ranges,  and  draw  the  diagram,  showing  regions  exposed  to  one  gun  only 
and  to  all  three  guns  (if  any). 


THE  CIRCLE 


93 


PROPOSITION  XIX.     THEOREM 
220.   Parallel  lines  intercept  equal  arcs  on  a  circle. 

M 


Vc 


N 

Given:  A  circle  and  a  pair  of  parallels  intercepting  two 
arcs. 

To  Prove :  The  intercepted  arcs  are  equal. 
There  may  be  three  cases  : 

I.  If  the  Us  are  a  tangent  (^1J?,  tangent  at  P)  and  a  secant 
(CD,  cutting  the  circle  at  E  and  F). 

Proof :  Draw  diameter  to  point  of   contact,  P. 

This  diameter  is  _L  to  AB  (203). 

PP'  is  also  _L  to  EF  (64). 

.-.  arc  EP  =  arc  l^P  (199). 

II.  If  the  Us  are  two  tangents,  points  of  contact  being  M 
and  N. 

Proof :  Suppose  a  secant  is  drawn  II  to  one  of  the  tangents, 
cutting  the  O  at  R  and  8. 

RS  will  be  II  to  the  other  tangent  (63). 

.-.  arc  MR  =  arc  M8  and  arc  RN  —  arc  SN  (I). 

Adding,                  arc  MRN  =  arc  MSN  (Ax,  2). 

III.  If  the  Us  are  two  secants,  one  cutting  the  O  at  A  and 
B,  the  other  at  C  and  D. 

Proof :  Suppose  a  tangent  is  drawn,  II  to  AB  and  touching 

the  O  at  P.     This  tangent  is  II  to  CD  (63). 

Arc  PC  =  arc  PD  and  arc  PA  =  arc  PB  (I). 

Subtracting,  arc  AC=  arc  BD  (Ax.  2). 

Q.E.D. 


94  BOOK  II.     PLANE   GEOMETRY 

221.    A  polygon  is  inscribed  ]  if   the  vertices   of   the  poly- 


in  a  circle,  or  a  circle  is  cir- 
cumscribed about  a  polygon 

A  polygon  is  circumscribed 
about  a  circle,  or  a  circle  is 


gon  are  in  the  circle,  and  its 
sides  are  chords. 

if   the  sides  of   the  polygon 


.,  are  all  tangent  to  the  circle. 

inscribed  m  a  polygon 

The  perimeter  of  a  figure  is  the  sum  of  all  its  bounding 

lines. 

EXERCISES  IN  DRAWING  CIRCLES 

1.  Draw  two  unequal  intersecting  circles.     Show  that  the  line  joining 
their  centers  is  less  than  the  sum  of  their  radii. 

2.  Draw  two  circles  externally  (not  tangent)  and  show  that  the  line 
joining  their  centers  is  greater  than  the  sum  of  their  radii. 

3.  Draw  two  circles  tangent  externally.     Discuss  these  lines  similarly. 

4.  Draw  two  circles  tangent  internally.     Discuss  these  lines  similarly. 
6.   Draw  two  circles  so  that  they  can  have  only  one  common  tangent. 

6.  Draw  two  circles  so  that  they  can  have  two  common  tangents. 

7.  Draw  two  circles  so  that  they  can  have  three  common  tangents. 

8.  Draw  two  circles  so  that  they  can  have  four  common  tangents. 

9.  Draw  two  circles  so  that  they  can  have  no  common  tangent. 

SUMMARY 

222.    The  following  is  a  summary  of  the  truths  relating  to 
magnitudes,  which  have  been  already  established  in  Book  II. 

I.   Arcs  are  equal  if  they  are : 

(1)  Intercepted  by  equal  central  angles. 

(2)  Subtended  by  equal  chords. 

(3)  Intercepted  by  parallel  lines. 

(4)  Halves  of  the  same  arc,  or  of  equal  arcs. 

II.   Lines  are  equal  if  they  are : 

(1)  Radii  of  the  same  or  of  equal  circles. 

(2)  Diameters  of  the  same  or  of  equal  circles. 

(3)  Chords  that  subtend  equal  arcs. 

(4)  Chords  that  are  equally  distant  from  the  center. 

(5)  Tangents  to  one  circle  from  the  same  point. 

III.   Unequal  arcs  and  unequal  chords  have  like  relations. 


ORIGINAL   EXERCISES  95 

ORIGINAL   EXERCISES 

1.  Show  that  an  inscribed  trapezoid  is  isosceles. 

2.  In  the  figure  of  242,  show  that  arc  DBX  =  arc  DB  +  arc  A  C. 

3.  In  the  figure  of  243,  show  that  arc  CA=arc  CMB-arc  CNB. 

4.  In  the  figure  of  244,  show  that  arc  CX  -  arc  CE  -  arc  BD. 

5.  In  the  figure  of  245,  show  that  arc  BX  =  arc  BE  —  arc  BD. 

6.  Show  that  the  perpendiculars  to  the  sides  of  a  circumscribed  poly- 
gon at  the  points  of  contact  meet  at  a  common  point. 

7.  Show  that  the  bisectors  of  the  angles  of  a  circumscribed  polygon 
meet  at  a  common  point. 

8.  If  two  circles  intersect  and  the  four  radii  are  drawn  to  the  points 
of  intersection,  prove  that  the  line  joining  the  centers  of  the  circles  bisects 
the  central  angles  formed  by  these  radii. 

9.  If  two  chords  of  a  circle  are  equal,  but  not  parallel,  and  their  mid- 
points are  joined  by  a  line,  prove  that  the  line  from  the  center  of  the  circle 
to  the  midpoint  of  the  other  line  is  perpendicular  to  it. 

10.  If  a  hexagon  is  circumscribed  about  a  circle,  prove  that  the  sum 
of  three  alternate  sides  equals  the  sum  of  the  other  three  sides. 

11.  Draw  two  circles  which  can  have  neither  a  common  chord  nor  a 
common  tangent. 

12.  Two  perpendicular  radii  are  prolonged  to  meet  a  tangent  to  a 
circle,  and  from  the  two  points  of  intersection  two  other  tangents  are 
drawn  to  this  circle.     Prove  that  these  two  tangents  are  parallel. 

(Hint.     Draw  radii  to  the  three  points  of  contact.) 

13.  If  from  the  midpoint  of  an  arc  perpendiculars  are  drawn  to  the 
radii  drawn  to  the  ends  of  the  arc,  prove  that  these  perpendiculars  are 
equal. 

14.  If  through  the  extremities  of  a  diameter  two  equal  chords  are 
drawn,  one  on  each  side  of  the  diameter,  prove  that  they  are  parallel. 

15.  Prove  the  theorem  of  210  by  the  accompanying 
figure. 

[Hint,  in  &AEK  show  that  two  sides  are  unequal, 
hence  two  A  are  unequal,  hence  two  angles  in  A  EKO 
are  unequal,  etc.] 

16.  Prove  the  theorem  of  211  by  this  figure,  and  a  method  similar  to 
that  employed  in  Ex.  15. 


96  BOOK   II.     PLANE   GEOMETRY 

KINDS   OF   QUANTITIES  —  MEASUREMENT 

223.  A  ratio  is  the  quotient  of  one  quantity  divided  by 
another  —  both  being  of  the  same  kind. 

224.  To  measure  a  quantity  is  to  find  the  number  of  times 
it  contains  another  quantity  of  the  same  kind,  called  the 
unit.     This  number  is  the  ratio  of  the  quantity  to  the  unit. 

225.  Two  quantities  are  called   commensurable  if  there 
exists  a  common  unit  of  measure  which  is  contained  in  each 
a  whole  (integral)  number  of  times. 

Two  quantities  are  called  incommensurable  if  there  does 
not  exist  a  common  unit  of  measure  which  is  contained  in 
each  a  whole  number  of  times. 

Thus,  $17  and  $35  are  commensurable,  but  $  17  and  $V35  are  incom- 
mensurable. Two  lines  18£  ft.  and  13  yd.  are  commensurable,  but  18£  ft. 
and  V13  yd.  are  incommensurable. 

226.  A  constant  quantity  is  a  quantity  the  value  of  which 
does  not  change  during  a  discussion.     A  constant  may  have 
only  one  value. 

A  variable  is  a  quantity  that  has  different  successive  values 
during  a  discussion.  It  may  have  an  unlimited  number  of 
values. 

227.  The  limit  of  a  variable  is  a  constant,  to  which  the 
variable  cannot  be  equal,  but  from  which  the  variable  can 
be  made  to  differ  by  less  than  any  mentionable  quantity. 

228.  Illustrative.     The  ratio  of  15  yd.  to  25  yd.  is  written  either  J$ 
or  15  -4-  25  and  is  equal  to  three  fifths.     If  we  state  that  a  son  is  two 
thirds  as  old  as  his  father,  we  mean  that  the  son's  age  divided  by  the 
father's,  equals  two  thirds.     A  ratio  is  a  fraction. 

The  statement  that  a  certain  distance  is  400  yd.  signifies  that  the  unit 
(the  yard),  if  applied  to  this  distance,  will  be  contained  exactly  400  times. 

Are  $  7.50  and  $3.58  commensurable  if  the  unit  is  $  1  ?  1  dime  ?  1  cent  ? 

Are  10  ft.  and  vT9  ft.  commensurable  ? 

The  height  of  a  steeple  is  a  constant;  the  length  of  its  shadow  made 
by  the  sun  is  a  variable.  The  distance  a  train  goes  varies  with  the  time 
it  travels.  Our  ages  are  variables.  The  length  of  a  standard  yard,  mile, 


MEASUREMENT  97 

or  meter,  etc.,  is  a  constant.     The  height  of  a  growing  plant  or  a  child  is 
a  variable. 

The  limit  of  a  variable  may  be  illustrated  by  considering  a  right  tri- 
angle ABC,  and  supposing  the  vertex  A 
to  move  farther  and  farther  from  the 
vertex  of  the  right  angle.  It.  is  evident 
that  the  hypotenuse  becomes  longer, 
that  AC  increases,  but  EC  remains  the 
same  length.  The  angle  A  decreases, 
the  angle  B  increases,  but  the  angle  C 
remains  constantly  a  right  angle.  If 
we  carry  vertex  A  toward  the  left  indefinitely,  the  ^  A  becomes  less  and 
less  but  cannot  become  zero.  [Because  then  there  could  be  no  A.] 

Hence  the  limit  of  the  decreasing  ZA  is  zero. 

Likewise,  the  Z  B  becomes  larger  and  larger  but  cannot  become  equal 
to  a  right  angle.  [Because  then  two  sides  of  the  triangle  would  be 
parallel,  which  is  impossible.]  But  it  may  be  made  as  nearly  equal  to 
a  right  angle  as  we  choose. 

Hence  the  limit  of  Z.  B  is  a  right  angle. 

To  these  limits  we  cannot  make  the  variables  equal,  but  from  these 
limits  we  can  make  them  differ  by  less  than  any  mentionable  angle,  how- 
ever small. 

The  following  supplies  another  illustration  of  the  limit  of  a  variable. 
The  sum  of  the  series  1  +  $  +  \  +  i  +  &  +  -h  +  -fa  +  T$T  +  etc->  win 
always  be  less  than  2,  no  matter  how  many  terms  are  collected.  But 
by  taking  more  and  more  terms  we  can  make  the  actual  difference 
between  this  sum  and  2  less  than  any  conceivable  fraction,  however 
small.  Hence  2  is  the  limit  of  the  sum  of  the  series.  The  limit  is  not 
3  or  4,  because  the  difference  between  the  sum  and  3  cannot  be  made 
less  than  any  assigned  fraction.  Neither  is  the  limit  1£.  (Why  not?) 
Similarly,  the  limit  of  the  value  of  .333333 ad  infinitum  is  %. 

Certain  variables  become  equal  to  a  fixed  magnitude ;  but  this  fixed 
magnitude  is  not  a  limit.  Thus,  the  length  of  the  shadow  of  a  tower 
really  becomes  equal  to  a  fixed  distance  (at  noon).  A  man's  age  really 
attains  to  a  definite  number  of  years  and  then  ceases  to  vary  (at  death). 

Hence  if  a  variable  approaches  a  constant,  and  the  differ- 
ence between  the  two  can  be  made  indefinitely  small  while 
the  variable  cannot  become  equal  to  the  constant,  the  con- 
stant is  the  limit  of  the  variable.  This  is  merely  another 
definition  of  a  limit. 


98  BOOK  II.     PLANE   GEOMETRY 

PROPOSITION  XX.     THEOREM  OF  LIMITS 

229.  If  two  variables  are  always  equal  and  each  approaches 
a  limit,  their  limits  are  equal. 

Given :  Two  variables  v  and  vf ;  v  always  =  v1 ;  also  v  ap- 
proaching the  limit  I ;  v'  approaching  the  limit  I'. 

To  Prove:  l  =  V. 

Proof:  v  is  always  =  to  v'  (Hyp.).  Hence  they  may  be 
considered  as  a  single  variable.  Now  a  single  variable  can 
approach  only  one  limit  (228).  Hence  1  =  1'.  Q.E.D. 

230.  (1)  Algebraic  principles  concerning  variables. 

If  v  is  a  variable  and  k  is  a  constant : 
I.    v  •+•  k  is  a  variable.  IV.    kv  is  a  variable. 

II.    v  —  k  is  a  variable.  V.    -  is  a  variable. 

k 

III.    k  ±  v  is  a  variable.  VI.    -  is  a  variable. 

v 
These  six  statements  are  obvious. 

(2)  Algebraic  principles  concerning  limits. 

If  v  is  a  variable  whose  limit  is  Z,  and  k  is  a  constant : 

I.    v  ±  k  will  approach  I  ±  k  as  a  limit. 
II.    k  ±  v  will  approach  k  ±  I  as  a  limit. 

III.  kv  will  approach  kl  as  a  limit. 

IV.  -  will  approach  -  as  a  limit. 

K  K 

k  k 

V.    -  will  approach  -  as  a  limit. 
v  I 

NOTE.  A  variable,  as  applied  to  Plane  Geometry,  is  not  added  to,  sub- 
tracted from,  multiplied  by,  or  divided  by  another  variable. 

Proofs:  I.  v  cannot  =  I  (227).     .•.  v  ±  k  cannot  =  I  ±  k. 

Also,  v  —  I  approaches  zero  (227) . 

.-.  (v  ±  k)  —  (/  ±  k)  approaches  zero.     (Because  it  reduces  to  v  —  /.) 

Hence  v  ±  k  approaches  I  ±  k  (227). 


MEASUREMENT   OF   ANGLES 


99 


II.   Demonstrated  similarly. 

III.   If   kv  =  klt  then  v  =  I  (Ax.  3).     But  this  is  impossible  (227). 
.•.  kv  cannot  =  kl. 

Also  v  —  I  approaches  zero  (227) . 
.•.  k(v  —  1)  or  kv  —  kl  approaches  zero. 
Therefore  kv  approaches  kl  (227). 
IV  and  V.     Demonstrated  similarly. 

PROPOSITION  XXI.     THEOREM 

231.    In  the  same  circle  (or  in  equal  circles)  the  ratio  of 
two  central  angles  is  equal  to  the  ratio  of  their  intercepted  arcs. 


Given : 
To  Prove  : 


O  o  =  O  c  ;  central 
O      arc  AB 


o  and  c  ;  arcs  AB  and  XT. 


C      arc  XY 

Proof:    I.  If  the  arcs  are  commensurable.     There  exists  a 
common  unit  of  measure  of  AB  and  XY  (225). 

Suppose  this  unit,  when  applied  to  the  arcs,  is  contained 
5  times  in  AB  and  7  times  in  XY. 

arc  AB  __  5 
"7 


(Ax.  3). 


arc  XY 

Draw  radii  to  the  several  points  of  division  of  the  arcs. 
O  is  divided  into  5  parts  and  Z  c  into  7  parts. 

These  12  parts  are  all  equal  (194). 

(Ax.  3). 


Zo 


Zo 
ZC 


5 

7 

arc  AB 
'  arc  XY 


(Ax.  1). 

Q.E.D. 


100  BOOK   II.     PLANE   GEOMETRY 

II.  If  the  arcs  are  incommensurable.  There  does  not  exist 
a  common  unit  (225).  Suppose  arc  AB  is  divided  into  equal 
parts  (any  number  of  them).  Apply  one  of  these  as  a  unit 
of  measure  to  arc  XT.  There  is  a  remainder  PT.  (Because 
AB  and  XT  are  incommensurable.) 


Draw  CP.     Now  --  =  ^.     (Case  I). 
Z  XCP     arc  XP 

Indefinitely  increase  the  number  of  subdivisions  of  arc  AB. 
Then  each  part,  that  is,  our  unit  or  divisor,  is  indefinitely 
decreased.     Hence  PF,   the  remainder,   is   indefinitely  de- 
creased.    (Because  the  remainder  <  the  divisor.) 
That  is,  arc  PT  approaches  zero  as  a  limit, 
and  Z.  PCT  approaches  zero  as  a  limit. 

.*.  arc  XP  approaches  arc  XT  as  a  limit  (227). 

and  Z.  XCP  approaches  Z.  XCT  as  a  limit  (227). 

Z  O  .  Z.O  ^'     -4. 

approaches  — —    -  as  a  limit 


ZXCP  ZXCT 

and  arcAB  approaches  arc  AB  as  a  limit. 
&TCXP  arc  XT 


(229) . 
XCT      arc  XT 


Ex.  1.   If  you  double  an  arc  do  you  double  its  central  angle  ?  its  chord  ? 

Ex.  2.  If  in  two  equal  circles,  an  arc  in  one  is  taken  three  times  as 
long  as  an  arc  in  the  other,  how  do  their  central  angles  compare?  Is 
there  any  similar  law  that  you  know,  applying  to  their  chords? 

Ex.  3.  Two  arcs  of  a  circle  contain  80°  and  120°  respectively.  What 
is  the  ratio  of  their  central  angles  ? 


MEASUREMENT   OF   ANGLES  101 

PROPOSITION  XXII.     THEOREM 

232.  A  central  angle  is  measured  by  its 
intercepted  arc. 

Given:    Oo;  Z.AOY;  arcJ.F. 

To  Prove  :  Z.  AOY  is  measured  by  the 
arc  AY,  that  is,  they  contain  the  same 
number  of  units. 

Proof  :   The  sum  of  all  A  about  O  =  4  rt.  A  =  360°      (47). 

If  this  O  is.  divided  into  360  equal  parts  and  radii  are 
drawn  to  the  several  points  of  division,  there  will  be  360 
equal  central  A  (194). 

Each  of  these  360  central  angles  will  be  a  degree  of  angle 

(20). 

Each  of  the  360  equal  arcs  is  called  a  degree  of  arc.  Take 
Z  AOT,  one  of  these  degrees  of  angle,  and  arc  AT,  one  of  the 


,.  rp,         Z.AOY      a.  /^OQ1\ 

degrees  of  arc.     Then  —  =  -  (231). 

Z  AOT      arc  AT 

Z  AOY  _  ^  AOY-L-  a  unit  =  the  number  of  units  in  Z  AOY 

(224). 

—  -  =  arc  AY-s-  a  unit  =  the  number  of  units  in  arc  AY 
we  AT 


.-.  the  number  of  units  in  Z  AOY=  the  number  of  units  in 
arc  AY  (Ax.  1). 

That  is,          /.  A  OY  is  measured  by  arc  AY. 

Q.E.D. 

233.  COROLLARY.    A  central  right  angle  intercepts  a  quadrant 
of  arc.     (Because  each  contains  90  units.) 

234.  COROLLARY.    A  right  angle  is  measured 
by  half  a  semicircle,  that  is,  by  a  quadrant. 

235.  An  angle  is  inscribed  in  a  segment  if 
its   vertex   is   on   the  arc  and   its   sides   are 
drawn  to  the  ends  of  the  arc  of  the  segment. 

Thus,  ABCD  is  a  segment  and  Z  ABD  is  inscribed  in  it. 


-102 


BOOK   II.     PLANE   GEOMETRY 


PROPOSITION  XXIII.     THEOREM 

236.    An  inscribed  angle  is  measured  by 
half  its  intercepted  arc. 

Given :  O  O  ;  inscribed  Z  A ;  arc  CD. 
To  Prove :  Z  A  is  measured  by  |  arc  CD. 

Proof :  I.     If  one  side  of  the  Z  is  a  diameter. 
Draw  radius  CO.     A  ^ioc  is  isosceles 


That  is, 
But 


•.  Z  COD  =  ^A  +  ^  A  = 
i  Z  COD  =  Z  ^ 

Z  COD  is  meas.  by  arc  CD 
*.  JZ  COD  is  meas.  by  \  arc  CD 
.-.  Z  4  is  meas.  by  J  arc  CD 


(?). 

(102). 

(?), 

(Ax.  6). 
(Ax.  3). 
(232). 
(Ax.  3). 
(Ax.  6), 


II.     If  the  center  is  within  the  angle. 
Z  CAX  is  measured  by  ^-  arc  CX 
Z  D4X  is  measured  by  J  arc 


.  Z  C^ID  is  measured  by  J  arc  CD 
III.  If  the  center  is  without  the  angle. 
Z  O4X  is  measured  by  J  arc  CX 
Z  D^JT  is  measured  by  |  arc  DX 
.  Z  CAD  is  measured  by  J  arc  CD 


Draw  diameter  AX. 

(i). 

(I).   Adding, 
(Ax.  2). 
Draw  diameter  AX. 

a)- 

(I).     Subtracting, 
(Ax.  2).        Q.E.D. 


MEASUREMENT   OF   ANGLES 


103 


237.  COROLLARY.    Angles  measured  by  half  the  same  arc, 
or  halves  of  equal  arcs,  are  equal. 

238.  COROLLARY.    In  the  same  circle  (or  in  equal  circles), 
equal  angles  are  measured  by  equal  arcs. 


PROPOSITION  XXIV.     THEOREM 

239.   All  angles  inscribed  in  the  same 
segment  are  equal. 

Given:   The  several  A  A  inscribed  in 
segment  BAG. 

To  Prove :  These  angles  all  equal. 

Proof :  Each  Z.  BAG  is  measured  by  J  arc  BC 
.-.  these  angles  are  all  equal. 


240.   COROLLARY.     All  angles  inscribed 
hi  a  semicircle  are  right  angles. 


Proof :  Each  Z  K  is  measured  by  J  a  semicircle 
.-.  each  Z  K  =  a  rt.  Z 


(236). 
(237). 

Q.E.D. 


(236). 
(234). 

Q.E.D. 


Historical  Note.  Thales,  a  Greek  from  Asia  Minor,  studied  geometry 
from  the  Egyptians  in  the  sixth  century  B.C.  He 
discovered  the  truth  of  240  as  well  as  a  number  of 
very  important  theorems.  For  example :  Book  I, 
Propositions  I,  II,  IV  and  XXXIV,  and  Book  III, 
Proposition  XX. 

Thales  was  one  of  the  "Seven  Wise  Men  of 
Greece,"  and  made  important  contributions  to 
astronomy  and  philosophy  as  well  as  to  geometry. 
He  regarded  water  as  the  principle  of  all  things. 


THALES 


104 


BOOK   II.     PLANE   GEOMETKY 


PROPOSITION  XXV.     THEOREM 

241.  The  angle  formed  by  a  tangent 

and  a  chord  is  measured  by  half  the 

intercepted  arc. 

Given :  O  o,  tangent  TN ;  chord  AP ; 

Z  TPA  ;   arc  PA. 

To  Prove :  Z  TPA   is   measured    by 

J  arc  PA. 

Proof :  Through  A  suppose  AX  drawn  II  to  TN. 
Now  Z  A  is  meas.  by  J  arc  PX 
But  Z  ^4  =  z  TP4 

Also  arc  PX  =  arc  P^4 

Substituting,  Z  TP^4  is  meas.  by  J  arc  P^l 


(236). 

(66). 

(220). 

(Ax.  6). 

Q.E.D. 


Ex.  1.  A  chord  divides  a  circle  into  two  arcs,  one  containing  100°,  the 
other,  260°.  An  angle  is  inscribed  in  each  segment.  How  many  degrees 
are  there  in  each  angle? 

Ex.  2.  In  a  circle,  an  inscribed  angle  and  a  central  angle  intercept  the 
same  arc,  which  contains  140°.  How  many  degrees  are  there  in  each  angle  ? 

Ex.  3.  A  chord  subtends  an  arc  of  74°.  How  many  degrees  are  there 
in  the  angle  between  the  chord  and  a  tangent  at  one  extremity  ? 

Ex.  4.  The  circumference  of  a  circle  is  divided  into  four  arcs,  40°,  70°, 
100°,  and  x.  Find  x  and  the  angles  of  the  quadrilateral  formed  by  the 
chords  of  these  arcs. 

Ex.  6.  In  a  segment  of  a  circle  whose  arc  contains  210°  is  inscribed 
an  angle.  How  many  degrees  are  there  in  this  angle? 

Ex.  6.  An  inscribed  angle  contains  35°.  How  many  degrees  are  there 
in  its  intercepted  arc  ? 

Ex.  7.  The  line  bisecting  an  inscribed  angle  bisects  also  its  inter- 
cepted arc. 

Ex.  8.   State  and  prove  the  converse  of  Ex.  7. 

Ex.  9.   The  line  bisecting  the  angle  between  a  tan- 
gent and  a  chord  bisects  the  intercepted  arc. 

Ex.  10.   State  and  prove  the  converse  of  Ex.  9.  Av 


MEASUREMENT   OF  ANGLES 


105 


Ex.  11.    The  angle  between  a  tangent  and  a  chord  is  half  the  angle 
between  the  radii  drawn  to  the  ends  of  the  chord. 

Ex.  12.  If  a  triangle  is  inscribed  in  a  circle  and  a 
tangent  is  drawn  at  one  of  the  vertices,  the  angles 
formed  between  the  tangent  and  the  sides  equals  the 
other  two  angles  of  the  triangle. 

Ex.  13.  By  the  figure  of  Ex.  12  prove  that  the  sum 
of  the  three  angles  of  a  triangle  equals  two  right  angles. 

Ex.  14.  Tf  one  pair  of  opposite  sides  of  an  inscribed 
quadrilateral  are  equal,  the  other  pair  are  parallel. 

Proof:    Draw   Js  BX,    CY;     arc   A B  =  arc   CD  (?). 
.-.  arc  ABC  =  arc  BCD  (Ax.  2). 

Hence  prove  rt.  &.ABX  and  DCY  equal. 

Ex.  16.    If  any  pair  of  diameters  is  drawn,  the  lines  joining  their 
extremities  (in  order)  form  a  rectangle. 

Ex.  16.  The  opposite  angles  of  an  inscribed  quadri- 
lateral are  supplementary. 

Ex.  17.  If  a  tangent  and  a  chord  are  parallel,  and  the 
chords  of  the  two  intercepted  arcs  are  drawn,  they  make 
equal  angles  with  the  tangent. 

Ex.  18.   The  circle  described  on  one  of  the  equal  sides  of  an  isosceles 
triangle  as  a  diameter  bisects  the  base. 

Proof :  Draw  line  EM.  The  A  .are  rt.  &  (?)  and 
congruent  (?). 

Ex.  19.  If  the  circle,  described  on  a  side  of  a  triangle 
as  diameter,  bisects  another  side,  the  triangle  is  isosceles. 

Ex.  20.    All  angles  that  are  inscribed  in  a  segment  greater  than  a 
semicircle  are  acute,  and  all  angles  inscribed  in  a  segment 
less  than  a  semicircle  are  obtuse. 

Ex.  21.    An  inscribed  parallelogram  is  a  rectangle. 
Prove  arc  ABC  =  arc  A  DC,  etc. 

Ex.  22.    The  diagonal  of  an  inscribed  rectangle  is  a  diameter 

Ex.  23.  A  circle  described  on  the  hypotenuse  of 
a  right  triangle  as  a  diameter  passes  through  the 
vertices  of  all  the  right  triangles  having  the  same 
hypotenuse. 

ROBBINS'S   NEW    PLANE    GEOM. — 8 


10(3 


BOOK   II.     PLANE   GEOMETRY 


Ex.  24.    If  from  one  end  of  a  diameter  a  chord  is  drawn,  a  perpen- 
dicular   to    it    drawn    from   the    other   end    of    the 
diameter  intersects  the  first  chord  on    the  circum- 
ference. 

Ex.  25.   If  two  circles  intersect  and  a  diameter  is 
drawn  in  each  circle  through  one  of  the  points  of 
intersection,  the  line  joining  the  ends  of  these  diameters 
the  other  point  of  intersection.     [Draw  chord  ABJ] 

Ex.  26.  If  a  tangent  is  drawn  at  one  end  of  a  chord, 
the  midpoint  of  the  intercepted  arc  is  equally  distant 
from  the  chord  and  the  tangent. 

[Draw  chord  A  M  and  prove  the  rt.  A  congruent.] 

Ex.  27.  If  two  circles  are  tangent  at  A  and  a  common 
tangent  touches  them  at  B  and  C,  the  angle  BA  C  is  a 
right  angle.  [Draw  tangent  at  A.  Use  206,  240.] 

Ex.  28.  The  bisectors  of  all  the  angles  inscribed  in 
the  same  segment  of  a  circle  pass  through  a  common  point 

Ex.  29.  If  two  circles  intersect  and  a  line  is 
drawn  through  each  point  of  intersection  terminat- 
ing in  the  circles,  the  chords  joining  these  extremi- 
ties are  parallel.  [Draw  RS.  Z  A  is  supp.  of 
^RSC  (?).  Finally  use  73.] 

Ex.  30.  A  circle  described  on  the  radius  of  another 
circle  as  diameter  bisects  all  chords  of  the  larger  circle 
drawn  from  their  point  of  contact. 

To  Prove:    AB  is  bisected   at  C. 

Proof:    Draw  chord  OC.     (Use  240,  200). 

Ex.  31.  If  two  equal  chords  intersect  within  a  circle,  the  segments  of 
one  are  equal  to  the  segments  of  the  other,  each  to  each. 

Ex.  32.  Prove  Proposition  XXV  by  drawing  a  diameter  to  the  point 
of  tangency,  instead  of  a  chord  parallel  to  the  tangent. 

Ex.  33.  If  in  figure  of  Ex.  25  above,  line  CD  met  the  two  circles  at  M 
and  N  instead  of  at  a  single  point  B,  what  could  be  said  of  the  lines 
AM  and  AN? 

Ex.  34.  If  a  circle  is  divided  into  four  equal  arcs  and  if  chords  of  these 
arcs  are  drawn,  the  inscribed  figure  is  a  square. 


MEASUREMENT   OF   ANGLES  107 

PROPOSITION  XXVI.     THEOREM 

242.  The  angle  formed  by  two  chords  intersecting  within  the 
circle  is  measured  by  half  the  sum  of  the  intercepted  arcs. 

(The  arcs  are  those  intercepted  by  the 
given  angle  and  by  its  vertical  angle.) 

Given:  Chords  AB  and  CD  intersect- 
ing at  P  ;  Z  APC ;  arcs  AC  and  DB. 

To  Prove:    Z  APC   is    measured     by 
J  (arc  AC+  arc  DB). 

Proof:   Suppose  CX  drawn  through  c  II  to  AB. 

Now        Z  c  is  measured  by  ^  arc  DX  (236). 

That  is,  Z  C  is  measured  by  J  (arc  BX  +  arc  DB). 

But  Z  c  =  Z  ^IPO  (66). 

Also  arc  BX  =  arc  AC  (220). 

.-.  Z.4PO  is  meas.  by  J  (arc  ^1(7  -f-  arc  DB)    (Ax.  6). 

Q.E.D. 

PROPOSITION  XXVII.     THEOREM 

243.  The  angle  formed  by  two  tangents  is  measured  by 
hah5  the  difference  of  the  inter- 
cepted arcs. 

Given:  The  two  tangents  AC 
and  AB ;  Z  A ;  arcs  CMB  and 
CNB.  /  N\  /M 

To  Prove :  Z  ^4  is  measured  by 
|  (arc  CMB  —  arc  CT.B). 

Proof :  Suppose  CX  drawn  II  to  AB. 

Now       Z  DOT  is  meas.  by  J  arc  CX  (241). 

That  is,  Z  DCX  is  meas.  by  |  (arc  CMB  —  arc  .BJT). 

But  ZDCX  =  ZA  (67). 

Also  arc  BX  —  arc  OZV5  (220). 

.  •.  Z  4  is  meas.  by  £  (arc  C3f£  —  arc  ora)     (Ax.  6). 

Q.E.D. 


108 


BOOK   11.     PLANE   GEOMETRY 


PROPOSITION  XXVIII.     THEOREM 

244.  The  angle  formed  by  two 
secants  which  intersect  without  the 
circle  is  measured  by  half  the  differ- 
ence of  the  intercepted  arcs. 

Given:  (?). 

To  Prove:  (?). 

Proof :  Suppose  BX  drawn.     Where  ?     How  ? 
Z  CBX  is  meas.  by  J  arc  CX 

That  is,  ^  CBX  is  meas.  by  |  (arc  CE  —  arc  XE) 

But  Z  CBX  —  Z  A 

And  arc  XE  =  arc  BD 

Substituting,  Z  A  is  meas.  by  £  (arc  CE  —  arc  BD) 


CO. 
(67). 

CO- 
(Ax.  6). 

Q.E.D. 


PROPOSITION  XXIX.     THEOREM 

245.  The  angle  formed  by  a  tangent 
and  a  secant  which  intersect  without 
the  circle  is  measured  by  half  the  dif- 
ference of  the  intercepted  arcs. 

Given:  (?). 

To  Prove:  (?). 

Proof :  Suppose  BX  drawn,  etc. 

Z  CBX  is  meas.  by  \  arc  BX  (241). 

That  is,  Z  CBX  is  meas.  by  J  (arc  BXE—  arc  XE). 

But  Z.CBX=^A  (?). 

And  arc  XE  —  arc  BD  (?). 

Substituting,  Z  A  is  meas.  by  J  (arc  BXE—  arc  .BD)      (?). 

Q.E.D. 

Ex.  1.  Where  is  the  vertex  of  an  angle  that  is  measured  by  one  arc? 
by  half  an  arc?  by  half  the  sum  of  two  arcs?  by  half  the  difference  of 
two  arcs? 

Ex.  2.   State  these  truths  all  in  a  single  theorem  of  your  own. 


ORIGINAL   EXERCISES  109 

ORIGINAL  EXERCISES 

1.  The  arcs  intercepted  by  two  secants  intersecting  without  a  circle 
contain  20°  and  140°  respectively.     How  many  degrees  are  there  in  the 
angle  formed  by  the  secants? 

2.  If  in  Ex.  1  the  intersecting  lines  were  chords,  how  many  degrees 
would  there  be  in  their  angle  ? 

3.  One  of  the  arcs  intercepted  by  two  intersecting  tangents  is  72°. 
Find  the  angle  formed  by  the  tangents. 

4.  Two  intersecting  chords  intercept  opposite  arcs  of  28°  and  80°. 
How  many  degrees  are  there  in  the  angle  formed  by  the  chords? 

6.   The  angle  between  a  tangent  and  a  chord  contains  27°.     How 
many  degrees  are  there  in  the  intercepted  arc  ? 

6.  The  angle  between  two  chords  is  30° ;  one  of  the  arcs  intercepted 
is  40°.    Find  the  other  arc.     [Denote  the  arc  by  #.] 

7.  If  in  figure  of  241,  arc  AP  contains  124°,  how  many  degrees  are 
there  in  Z  TPA  ?  in  Z1  NPA  ?  in  arc  AX  ? 

8.  If  in  figure  of  242,  arc  A  C  is  85°,  Z  A  PC  is  47°,  find  arc  DB. 

9.  If  the  arcs  intercepted  by  two  tangents  contain  80°  and  280°,  find 
the  angle  formed  by  the  tangents. 

10.  If  the  arcs  intercepted  by  two  secants  contain  35°  and  185°,  find 
the  angle  formed  by  the  secants. 

11.  If  in  figure  of  243,  arc  CB  is  135°,  find  the  angle  A. 

12.  If  in  figure  of  244,  angle  A  =  42°  and  arc  BD  =  70°,  find  arc  CE. 

13.  If  in  figure  of  245,  angle  A  =  18°,  arc  BXE  =  190°,  find  arc  BD. 

14.  If  the  angle  between  two  tangents  is  80°,  find  the  number  of 
degrees  in  each  intercepted  arc.     [Denote  the  arcs  by  #  and  360°  —  x.~\ 

15.  Three  of  the  intercepted  arcs  of  a  circumscribed  quadrilateral  are 
68°,  98°,  114°.     Find  the  angles  of  the  quadrilateral.     If   the  chords 
are  drawn  connecting  (in  order)  the  four  points  of  contact,  find  the 
angles  of  this  inscribed  quadrilateral.     Also  find  the  angles  between  the 
diagonals  of  the  two  quadrilaterals. 

16.  If  the  angle  between  two  tangents  to  a  circle  is  40°,  find  the  other 
angles  of  the  triangle  formed  by  drawing  the  chord  joining  the  points 
of  contact. 


110 


BOOK   II.     PLANE   GEOMETKY 


17.  The  circumference  of  a  circle  is  divided 
into  four    arcs,   three  of   which   are,   RS  =  62°, 
ST  =  142°,  TU  =  98°.     Find  : 

(1)  Arc  UR. 

(2)  The  three  angles  at  R  ;  at  S\  at  T;  at  U. 

(3)  The  angles  A,  B,  C,  D  of  circumscribed 
quadrilateral. 

(4)  The  angles  between  the  diagonals  R  T  and  SU. 

(5)  The  angle  between  RL7  and  ST1  at  their  point  of  intersection 
(if  produced). 

(6)  The  angle  between  RS  and  TU  at  their  intersection. 

(7)  The  angle  between  AD  and  BC  at  their  intersection. 

(8)  The  angle  between  AB  and  DC  at  their  intersection. 

(9)  The  angle  between  RS  and  DC  at  their  intersection. 
(10)  The  angle  between  AD  and  S^at  their  intersection. 

18.  If  in  the  figure  of  Ex.  17,  Z  A  =96°;  ZB  =  112°  ;  and  /.  C  =  68°, 
find  the  angles  of  the  quadrilateral  RSTU.    [Denote  arc  RU  by  x.    .•.  in 
A  ARUy  96°  +  i  x  +  \  x  =  180°.     .-.  x  =  etc.] 

19.  If  two  circles  are  tangent  externally  and 
any  line  through  their  point  of  contact  intersects 
them  at  B  and  C,  the  tangents  at  B  and  C  are  par- 
allel. [Draw  common  tangent  at  A.  Prove  :  Z  A  CT 


20.  Prove  the  same  theorem  if  the  circles  are 
tangent  internally. 

21.  If  two  circles   are   tangent   externally   and  any  line   is   drawn 
through  their  point  of  contact  terminating  in  the  circles,  the  two  diame- 
ters drawn  to  the  extremities  are  parallel. 

22.  Prove  the  same  theorem  if  the  circles  are  tangent  internally. 

23.  If  two  circles  are  tangent  externally  and 
any  two  lines  are  drawn  through  their  point  of 
contact  intersecting  the  circles,  the  chords  joining 
these  points  of  intersection  are  parallel. 

[Draw  common  tangent  at  0.  Prove  :  Z  C  =  /.  D.~] 

24.  Prove  the  same  theorem  if  the  circles  are 
tangent  internally. 

25.  Prove  the  theorem  of  242  by  drawing  AD  instead  of  CX,  and 
using  Z.APC  as  an  exterior  angle  of  A  APD. 

26.  Prove  the  theorem  of    213  by  drawing  BC  and  using  /.DCB 
as  an  exterior  angle  of  A  ABC. 


ORIGINAL   EXERCISES  111 

27.  Prove  the  theorem  of  244  by  drawing  CD  and  using  angle  CDE 
as  an  exterior  angle  of  triangle  A  CD. 

28.  Prove  the  same  theorem  by  drawing  BE. 

29.  Prove  the  theorem  of  245  by  drawing  BE. 

30.  If  the  opposite  angles  of  a  quadrilateral  are  supplementary,  a 
circle  can  be  drawn  circumscribing  it.  p 

To  Prove  :   A  O  can  be  drawn  through  A,  B,  C,  P. 

Proof:  A  O  can  be  drawn  through  A,  B,  C  (?).  It 
is  required  to  prove  that  it  will  contain  point  P. 
/_  P  +  Z  B  are  supp.  (?)  .-.  they  must  be  meas.  by  half  the 
entire  circle.  Z  B  is  meas.  by  |  arc  ADC  (?).  Hence  /.P  is  meas. 
by  \  arc  ABC.  If  ZP  is  within  or  without  the  circle,  it  is  not  meas.  by 
%  arc  ABC.  (Why  not?) 

31.  The  circle  described  on  the  side  of  a  square,  or  of  a  rhombus, 
as  a  diameter  passes  through  the  point  of  intersection  of  the  diagonals. 
[Use  135,  141.] 

32.  The  line  joining  the  vertex  of  the  right  angle 
of  a  right  triangle  to  the  point  of  intersection  of  the 
diagonals  of  the  square  constructed  upon  the  hypotenuse 
as  a  side,  bisects  the  right  angle  of  the  triangle. 


Proof :   Describe  a  O  upon  the  hypotenuse  as  diameter  and  use  141, 
196,  237. 

33.  If  two  secants,  PAR  and  PCD,  meet  a  circle  at  A,  B,  C,  and 
D,  respectively,  the  triangles  PBC  and  PAD  are  mutually  equiangular. 

34.  If  PAB  is  a  secant  and  PM  is  a  tangent 
to  a  circle  from  P,  the  triangles  PAM  and  PBM 
are  mutually  equiangular. 

35.  If  two  equal  chords  intersect  within  a  circle, 

(1)  One  pair  of  intercepted  arcs  are  equal. 

(2)  Corresponding  parts  of  the  chords  are  equal. 

(3)  The   lines  joining   their  extremities    (in   order) 
form  an  isosceles  trapezoid. 

(4)  The  radius  drawn  to  their  intersection  bisects  their  angle. 


112 


BOOK   II.    PLANE   GEOMETRY 


36.  If    a   secant  intersects  a  circle  at  D  and  E,  PC 
is  a  parallel  chord,  and  PR  a  tangent  at  P  meeting  the 
secant  at  R,  the  triangles  PCD  and  PRD  are  mutually 
equiangular.    [/.  R  and  /.  CDP  are  measured  by  £  arc  PC. 
(Explain.)     Etc.] 

37.  If  a  circle  is  described  upon  one  leg  of  a  right  triangle  as  diameter 
and  a  tangent  is  drawn  at  the  point  of  its  intersec- 
tion  with   the   hypotenuse,    this    tangent   bisects  the 

other  leg. 

[Draw  OP  and  OD.  CD  is  tangent  (?).  OD  bi- 
sects arc  PC  (207).  ^COD=ZA  (237).  .-.  OD  is 
II  to  AB  (?).  Etc.] 

38.  If  an   equilateral  triangle  ABC  is  inscribed  in  a 
circle  and  P  is  any  point  of   arc  AC,  AP  +  PC  =  BP. 
[Take  PN  =  PA  ;   draw  AN.      A  ANP   is  equilateral. 
(Explain.)     AANB  =  &APC(?).     Etc.] 

39.  If    two  circles   are   tangent  internally  at  C,  and 
a  chord  AB  of  the  larger  circle  is  tangent  to  the  less 
circle  at  M,  the  line  CM  bisects  the  angle  A  CB. 

[Draw  tangent  CX  and  chord  RS.     (Explain.) 


.-.  AB  is  ||  to  RS  (?).     Etc.]  V 

40.  If  two  circles  intersect  at  A  and  C  and  lines 
are  drawn  from  any  point  P,  in  one  circle,  through 
A  and  C  terminating  in  the  other  at  points  B  and 
Z>,  chord  BD  will  be  of  constant  length  for  all  posi- 
tions of  point  P. 

[Draw  EC.     Prove  /.BCD,  the  ext.  Z  of  APBC, 
=  a  constant.     Etc.] 

41.  The  perpendiculars  from  the  vertices  of 
a  triangle  to  the  opposite  sides  are  the  bisectors 
of  the  angles  of  the  triangle  formed  by  joining 
the  feet  of  these  perpendiculars. 

To  Prove  :  BS  bisects  Z.  RST,  etc. 

Proof:  If  a  circle  is  described  on  AO  as  diarn.,  it  will  pass  through  T 
and  S  (141).  If  a  circle  is  described  on  OC  as  diam.,  it  will  pass 
through  R  and  ,S  (?).  .-.  ZBAR  =  BST  (?)  ;  and  £  BCT  =  Z.  BSR  (?). 
But  Z.  BAR  =  Z BCT.  (Each  is  the  comp.  of  £  ABC.)  .-.  Etc. 


ORIGINAL  EXERCISES  113 

42.  If  ABC  is  a  triangle  inscribed  in  a   circle,   BD 
is  the  bisector  of  angle  ABC,  meeting  A  C  at  O  and  the 
circle  at  D,  the  triangles  A  OB  and  COD  are  mutually 
equiangular.      Also    triangles  BOC    and    AOD.     Also 
triangles  BOC  and  ABD.    Also  triangles  A  OD  and 
Also  triangles  BCD  and  COZ). 

43.  If  two  circles  intersect  at  A  and  J3,  and  from  P,  any  point  on 
one  of  them,  lines  A  P  and  BP  are  drawn  cutting  the  other  circle  again 
at  C  and  D  respectively,  CD  is  parallel  to  the  tangent  at  P. 

44.  If  two  circles  intersect  at  A   and  B,  and  through  B  a  line  is 
drawn    meeting  the   circles  at  R  and  5  respectively,  the  angle  RAS 
is  constant  for  all  positions  of  the  line  RS. 

[Prove  /.  R  +  Z.  S  is  constant.     .-,  Z.  RAS  is  also  constant.] 

45.  Two  circles  intersect  at  A,  and  through  A  any  secant  is  drawn  meet- 
ing the  circles  again  at  M  and  N.     Prove  that  the  tangents  at  M  and  N 
meet  at  an  angle  which  remains  constant  for  all  positions  of  the  secant. 

[Prove  the  angle  between  these  tangents  equal  to  the  angle  between 
the  tangents  to  the  circles  at  A .] 

46.  Two   equal   circles   intersect   at  A    and  B,  and  through  A  any 
straight  line  MAN  is  drawn,  meeting  the  circles  at  M  and  N  respectively. 
Prove  chord  BM  =  chord  BN. 

47.  If    the  midpoint  of   the  arc  subtended  by  any  chord  is  joined 
to  the  extremities  of  any  other  chord, 

(1)  The  triangles  formed  are  mutually  equiangular.  (2)  The  op- 
posite angles  of  the  quadrilateral  thus  formed  are  supplementary. 

48.  Two  circles  meet  at  A   and  B  and  a  tangent  to  each  circle  is 
drawn  at  A,  meeting  the  circles  at  R  and  S  respectively.    Prove  that  the 
triangles  ABR  and  ABS  are  mutually  equiangular. 

49.  Two  chords  intersecting  within   a  circle  divide  the  circumfer- 
ence into  parts  that  bear  the  relation  to  each  other  of  1,  2,  3,  4.     Find 
the  angles  made  by  the  chords.     [Denote  the  arcs  by  ar,  2  x,  3  x,  4  #.] 

50.  If  A  BCD  is  an  inscribed  quadrilateral,  AB  and  DC  produced 
to  meet  at  E,  AD  and  EC  produced  to  meet  at  Ft  the  bisectors  of  angles 
E  and  F  are  perpendicular. 

[The  difference  of  one  pair  of  arcs  =  difference  of  a  second  pair ;  the 
difference  of  a  third  pair  =  difference  of  a  fourth  pair.  (Explain.) 
Transpose  negative  terms  and  add  correctly,  rioting  that  the  sum  of  4 
arcs  =  sum  of  4  others,  and  hence  =  180°.  Half  the  sum  of  these  4  arcs 
measures  the  angle  between  the  bisectors.  (Explain.)  Etc.] 


114  BOOK   II.     PLANE   GEOMETRY 

LOCI 

246.  The  locus  of  a  point  is  the  series  of  positions  the 
point  must  occupy  in  order  that  it  may  satisfy  a  given  con- 
dition.    It  is  the  path  of  a  point  whose  positions  are  limited 
or  denned  by  a  given  condition,  or  given  conditions. 

247.  Explanatory.     I.    If  a  point  is  moving  so  that  it  is 
always  one  inch  from  a  given  point,  the  moving  point  may 
occupy  any  position  in  a  circle  whose  center  is  the  fixed  point 
and  whose  radius  is  one  inch.      Furthermore,  this  moving 
point  cannot  occupy  any  position  outside  of  the  circle,  or  its 
position  will  not  fulfill  the  given  condition. 

THEOREM.  The  locus  of  points  in  a  plane  a  given  dis- 
tance from  a  given  point  is  a  circle  the  center  of  which  is 
the  given  point  and  the  radius  of  which  is  the  given  distance. 

II.  If  a  point  is  moving  so  that  it  is  always  equally  dis- 
tant from  the  ends  of  a  straight  line,  it  must  move  in  the 
perpendicular  bisector  of  the  line. 

THEOREM.  The  locus  of  points  equally  distant  from  the  ends 
of  a  line  is  the  perpendicular  bisector  of  the  line. 

Proof:  Every  point  in  the  _L  bisector  is  equally  distant 

from  the  ends  of  the  line.  (80.) 

No  point  without  this  _L  fulfills  that  condition.  (81.) 

.-.  the  JL  bisector  is  the  locus.  (246.) 

III.  THEOREM.     The  locus  of  points  equally  distant  from 
the  sides  of  an  angle  is  the  bisector  of  the  angle. 

Proof :  Any  point  within  the  bisector  of  an  angle  is  equally 
distant  from  the  sides.  (94.) 

Any  point  equally  distant  from  the  sides  of  an  angle  lies 
in  the  bisector.  (95.) 

Hence  all  the  points  in  the  bisector  fulfill  the  condition 
and  there  are  no  other  points  that  fulfill  it. 

"That  is,  the  bisector  is  the  locus,  etc.  Q.E.D. 


ORIGINAL   EXERCISES   ON   LOCI  115 

IV.  The  method  of  proving  that  a  certain  line  or  a  group 
of  lines  is  the  locus  of  points  satisfying  a  given  condition, 
consists  in  proving  that  every  point  in  the  line  fulfills  the 
given  requirement,  and  that  there  is  no  other  point  that  ful- 
fills it.  In  the  above  illustrations  it  is  evident  that  every 
point  in  the  lines  that  were  called  the  "locus,"  fulfilled 
the  conditions  of  the  case.  It  is  evident  also  that  there  is 
no  point  outside  these  "loci"  that  does  so  fulfill  the  con- 
ditions. That  is,  these  uloci"  contain  all  the  points  de- 
scribed. 

ORIGINAL  EXERCISES   ON   LOCI 

1.  What  is  the  locus  of  a  point  so  moving  that  it  is  always  two  feet 
away  from  a  given  line  ? 

2.  What  is  the  locus  of  a  point  so  moving  that  it  is  always  equally 
distant  from  two  parallel  lines  ? 

3.  What  is  the  locus  of  points  equally  distant  from  two  given  fixed 
points  ? 

4.  If  all  the  radii  of  a  circle  were  drawn,  what  would  be  the  locus  of 
their  midpoints  V 

5.  If  all  possible  lines  were  drawn  from  a  vertex  of  a  triangle  and 
terminating  in  the  opposite  side,  what  would  be  the  locus  of  their  mid- 
points ? 

6.  What  is  the  locus  of  the  midpoints  of  a  series  of  parallel  chords 
in  a  circle?    Prove. 

7.  What  is  the  locus  of  the  midpoints  of  all  chords  of  the  same 
length  in  a  given  circle?    Prove. 

8.  What  is  the  locus  of  all  points  from  which  two  equal  tangents 
can  be  drawn  to  two  circles  which  are  tangent  to  each  other? 

9.  What  is  the  locus  of  all  points  at  a  given  distance  from  a  given 
circumference  ?     Discuss  if  the  distance  is  >  radius.     If  it  is  less. 

10.  What  is  the  locus  of  the  vertices  of  the  right  angles  of  all  the 
right  triangles  that  can  be  constructed  on  a  given  hypotenuse  ?     Prove. 

11.  What  is  the  locus  of  the  vertices  of  all  the  triangles  which  have  a 
given  acute  angle  (at  that  vertex)  and  have  a  given  base  ?    Prove. 


116  BOOK   II.     PLANE   GEOMETRY 

12.  A  line  of  given  length  moves  so  that  its  ends  are  in  two  perpen- 
dicular lines.     What  is  the  locus  of  its  midpoint?     Prove. 

[Suppose  AB  represents   one  of   the  positions   of       J 
the  moving  line.     Draw  OP  to  its  midpoint.     In  all 
the  positions  of  AB,  OP  =  |  A B  -  a  constant  (141). 

.*.  P  is  always  a  fixed  distance  from  0.     Etc.] 

13.  What  is  the  locus  of  the  midpoints  of  all  the  chords  that  can  be 
drawn  through  a  fixed  point  on  a  given  circle? 

[Suppose  AB  represents  one  of  the  chords  from  B 
in  circle  0,  with  radius  OB ;  and  P  is  the  midpoint  of 
A  B.     Draw  OP.     Z  P  is  a  rt.  Z  (?).     That  'is,  where- 
ever  the  chord  may  be  drawn,  Z  P  is  a  rt.  Z. 
.'.  locus  of  P  is,  etc.] 

14.  A  definite  line  which  is  always  parallel  to  a  given  line  moves  so 
that  one  of  its  extremities  is  on  a  given  circle.     Find  the  locus  of  the 
other  extremity. 

[Suppose  CP  represents  one  position  of          ^ ^xC p 

the  moving  line    CP.      Draw   OQ  =  and   II 

to   CP  from   center    0.     Join  OC  and  PQ. 

Wherever  CP  is,  this  figure  is  a  O  (?).     Its 

sides  are  of  constant  length  (?).     That  is,  P  ^ 

is  always  a  fixed  distance  from  Q,  etc.]  A B 

15.  What  is  the  locus  of  the  centers  of  all  circles  tangent  to  a  given 
line  at  a  given  point  ?  to  a  given  circle  at  a  given  point  ? 

16.  A  parallelogram,  ABCD,  is  hinged  at  the  vertices,  and  AB  only  is 
fixed  in  position.     What  is  the  locus  of  vertex  C?    of  vertex  D't  of  the 
midpoint  of  BC1   of  the  midpoint  of  CDt 

248.  Heretofore  only  a  few  of  the  simplest  exercises  in 
construction  have  been  given*  (pages  8-12),  and  formal  proofs 
of  these  were  not  required. 

The  following  methods  for  constructing  lines  are  given 
so  that  mathematical  precision  may  be  employed  in  drawing 
accurate  diagrams  of  a  complex  nature.  No  construction 
is  considered  valid  unless  a  proof  of  its  correctness  can  be 
given. 

The  pupil  should  be  familiar  with  the  use  of  the  ruler 
and  compasses. 


CONSTRUCTION   PROBLEMS  117 

CONSTRUCTION   PROBLEMS 

249.  A  geometrical  construction  is   a   diagram   made   of 
points  and  lines. 

250.  A  geometrical  problem  is  the  statement  of  a  required 
construction.     Thus,  "  It  is  required  to  bisect  a  line "  is  a 
problem.     A  problem  is  sometimes  defined  as  "  a  question  to 
be   solved"  and   includes  other  varieties  besides  those  in- 
volved in  geometry. 

251.  The  word  proposition  is  used  to  include  both  theorem 
and  problem. 

252.  The  complete  solution  of  a  problem  consists  of  five 
parts: 

I.    The  Given  data  are  to  be  described. 
II.    The  Required  construction  is  to  be  stated. 

III.  The  Construction  is  to  be  outlined. 

This  part  usually  contains  the  verb  only  in  the  imperative. 
The  only  limitation  in  this  part  of  the  process  is  that  every 
construction  demanded  shall  have  been  shown  possible  by 
previous  constructions  or  postulates.  (See  32,  33,  186.) 

IV.  The  Statement  that   the   required   construction   has 
been  completed. 

V.    The  Proof  of  this  declaration. 

Frequently  a  discussion  of  ambiguous  or  impossible  in- 
stances is  necessary. 

253.  NOTES.  (1)  A  straight  line  is  determined   by  two 
points. 

(2)  A  circle  is  determined  by  three  points. 

(3)  A  circle  is  determined  by  its  center  and  its  radius. 
Whenever  a  circumference,  or  even  an  arc,  is  to  be  drawn, 
it  is  essential   that  the  center  and  the  radius  be  mentioned. 

(4)  "Q.E.F.  "  =  Quod  erat  faciendum  —  "which  was  to  be 
done."     These  letters  follow  the  statement  that  the  construc- 
tion which  was  required  has  been  accomplished. 


118 


BOOK   II.     PLANE   GEOMETRY 


PROPOSITION  XXX.     PROBLEM 

254.    To  bisect  a  given  line. 

Given :  The  definite  line  AB. 

Required :    To  bisect  AB. 

Construction:    Using   A    and    B    as 
centers    and    one    radius,    sufficiently     A  | 

long  to  make  the  circumferences  in- 
tersect, describe  two  arcs  meeting  at 
B  and  T.  Draw  ET  meeting  AB  at  M. 

Statement :  Point  M  bisects  AB. 

Proof :  B  is  equally  distant  from  A  and  B 

T  is  equally  distant  from  A  and  B 

Hence  ET  is  the  _L  bisector  of  AB 

That  is,  M  bisects  AB. 


M 


;<r 


Q.E.F. 

(188). 

(?)• 

(83). 

Q.E.D. 


PROPOSITION  XXXI.    PROBLEM 
255.    PROBLEM.     To  bisect  a  given  arc. 
Given:  Arc  AB  whose  center  is  O. 
Required:  To  bisect  arc  AB. 
Construction :    Draw  chord  AB.     Us- 
ing A  and  B  as  centers  and  any  suffi- 
cient radius,  describe  arcs  meeting  at 
C.     Draw  OC  cutting  arc  AB  at  M. 
Statement :  The  point  M  bisects  arc  AB. 
Proof :  O  and  C  are  each  equally  distant  from  A  and  B 
.'.  OC  is  -the  _L  bisector  of  chord  AB 
.'.  M  bisects  arc  AB 


Q.E.F. 

(188), 

(83), 

(200). 

Q.E.D. 


Ex.  1.     Construct  the  supplement  of  a  given  angle. 
Ex.  2.     Divide  a  given  line  into  four  equal  parts. 
Ex.  3.     Divide  a  given  arc  into  four  equal  arcs. 


CONSTRUCTION   PROBLEMS 


119 


PROPOSITION  XXXII.     PROBLEM 
256.    To  bisect  a  given  angle. 
Given:  Z  LON. 
Required :  To  bisect  Z  LON. 
Construction:    Using   o   as   a  center 
and  any  radius,  draw  arc  AB,  cutting 
LO  at  A  and  NO  at  B.     Using  A  and  B 
as  centers   and   any  sufficient    radius, 
draw  two  arcs  intersecting  at  S.     Draw 
OS  meeting  arc  AB  at  M. 
Statement :  OS  bisects  Z  LON. 
Proof :  Draw  AS  and  BS.     In  A  AOS  and  BOS,  OS 
OA  =  OB  and  AS  =  BS 
.'.  A  AOS  ^  A  BOS 
.'.  Z  AOS  =  Z  BOS 


Q.E.F. 
OS   (?). 

(188). 


Q.E.D. 


;s 


PROPOSITION  XXXIII.     PROBLEM 

257.    At  a  fixed  point  in  a  straight  line  to  erect  a  perpen- 
dicular to  that  line. 

Given:    Line    AB     and    point    P    with- 
in it. 

Required:  To  erect  a  line  J_  to  AB  at  P. 

Construction:  Using  P  as  a  center     A-pt 

and  any  radius,  draw  arcs  meeting 

AB  at  C  and  D.      Using  C  and  D  as  centers  and  a  radius 

longer  than  before,  draw  arcs  meeting  at  S.     Draw  PS. 

Statement:  PS  is  J.  to  AB  at  P.  Q.E.F. 

Proof:  Points  P  and  S  are  each  equally  distant  from  C 

and  D.  (188). 

.-.  PS  is  the  -L  bisector  of  CD  (83). 

That  is,  PS  is  A.  to  AB.  Q.E.D. 


120  BOOK   II.    PLANE   GEOMETRY 


Another  Construction :  Using  any 
point  O,  without  AB,  as  center,  and 
OP  as  radius,  describe  a  circum- 
ference, cutting  AB  at  P  and  E. 
Draw  diameter  EOS.  Join  SP. 

Statement :  -SP  is  _L  to  AB  at  P.      A 

Q.E.F. 

Proof :  Segment  SPE  is  a  semicircle 


%  Z  SPEis  a  rt.  Z  (240). 

/.  SP  is  J_  tO  4£  (16).      Q.E.D. 


Ex.  1.   Construct  a  right  angle. 

Ex.  2.    Construct  an  angle  of  45°;  of  135°  ;  of  22°  30'  ;  of  67°  30'. 

Ex.  3.    Construct  the  complement  of  a  given  angle. 

Ex.  4.   Find  the  center  of  a  given  circle. 

Ex.  5.   Construct  the  second  acute  angle  of  a  rt.  A  if  one  is  known. 

Ex.  6.   Construct  a  chord  of  a  circle  if  its  midpoint  is  known. 

PROPOSITION  XXXIV.     PROBLEM 

258.   Through  a  point  without  a  line  to  draw  a  perpendicular 
to  that  line. 

Given  :  Line  AB  and  point  P  with- 
out  it. 


Required:  (?). 

Construction  :  Using  P  as  a  center    A      -M 


N 


and  any  sufficient  radius,  describe  an 

arc  intersecting  AB  at  M  and  N.  Using  M  and  N  as  centers 
and  any  sufficient  radius,  describe  arcs  intersecting  each 
other  at  C.  Draw  PC. 

Statement:  PC  is  _L  to  AB  from  P.  .  Q.E.F. 

Proof  :  P  and  C  are  each  equally  distant  from  M  and  N 

(188). 

.-.  PC  is  the  -L  bisector  of  MN  (83). 

That  is,  PC  is  J_  to  AB  from  P.  Q.E.D. 


CONSTRUCTION  PROBLEMS  121 

PROPOSITION  XXXV.     PROBLEM 

259.    At  a  given  point  in  a  given  line  to  construct  an  angle 
which  shall  be  equal  to  a  given  angle.  A 

Given:  Z  AOB  ;  point  P  in  line  CD. 

Required:    To   construct    at   P    an 
Z=to  Z  AOB. 

Construction:  Using  o  as  a  center 
with  any  radius,  describe  an  arc  cutting 
OA  at  E  and  OB  at  F.  Draw  chord  EF.  Using  P  as  a  center 
and  OE  as  a  radius,  describe  an  arc  cutting  CD  at  R.  Using 
R  as  a  center  and  chord  EF  as  a  radius,  describe  an  arc  cut- 
ting the  former  arc  at  X.  Draw  PX  and  chord  EX. 

Statement  :  Z  XPD  =  Z  AOB.  Q.E.F. 

Proof:     OE  =  PX  and  OF=  PR  and  EF=  XR  (188). 

.-.  A  OEF  ^  A  PXR  (?)  . 


Q.E.D. 


PROPOSITION  XXXVI.     PROBLEM 


260.   To  draw  a  line  through  a  given  point  parallel  to  a  given 
line. 

Given  :  Point  P  and  line  AB.  p\  -  x 

Required  :  To  draw  through  P,  a  line  \  _ 

\\toAB.  A       N~ 

Construction:  Draw  through  P  any  line  PN,  meeting  AB 
at  N. 

On  this  line,  at  P,  construct  Z  jvrpjr  =  to  Z  ANP       (259). 

Statement:  PX  is  II  to  AB.  Q.E.F. 

Proof:  Z^PX=Z^JVP  (Const.). 

.-.  PX  is  II  to  AB  (70). 

Q.E.D. 

KOBBINS'S    NEW   PLANE    GEOM.  —  9 


2  BOOK   II.     PLANE   GEOMETKY 

PROPOSITION  XXXVII.     PHOBLEM 
261.   To  divide  a  line  into  any  number  of  equal  parts. 
Given  :  Definite  line  AB. 

A      O       N      M      L       B 

Required:  To    divide    it    into    five 


equal  parts.  ''^...    /      I      j 

Construction:  Draw  through  A  any  E  "-•-,../      / 

other    line    AX.     On    this    take    any  '"£  '-,... 

length  AC  as  a  unit,  and  mark  off  on 
AX  five  of  these  units,  AC,  CD,  DE,  EF,  FG.     Draw  GB. 

Through  F,  E,  D,  C,  draw  II  to  GB,  lines  FL,  EM,  DN,  CO. 

Statement  :  Then  AO  =  ON  =  NM  =  ML  =  LB.  Q.E.F. 

Proof:  AC=CD=  DE=  EF=FG  (Const.). 

.'.  AO  =  ON  =  NM  =  ML  =  LB  (140). 

_  Q.E.D. 

Ex.  1.  Find  a  point  in  one  side  of  a  triangle  equally  distant  from 
the  other  sides. 

Ex.  2.  Construct  the  shortest  chord  that  can  be  drawn  through  a 
given  point  within  a  circle. 

Proof  :  Draw  any  other  chord  through  the  point,  etc. 

Ex.  3.  Draw  through  a  given  point  without  a  given  line,  another 
line  which  shall  make  a  given  angle  with  the  line.  ^ 

Construction:  Construct  an  Z  at  any  point  in 
the  given  line  =  to  given  Z,  etc. 

Ex.  4.  Construct  through  a  given  point  a  line 
which  makes  equal  angles  with  two  intersecting 
lines.  "^B 

Ex.  6.  Bisect  the  three  sides  of  a  triangle,  and  draw  the  medians. 
Do  they  meet  in  a  point? 

Ex.  6.  Draw  accurately  the  three  altitudes  of  a  triangle.  Do  they 
meet  in  a  point? 

Ex.  7.  Draw  accurately  the  three  perpendicular  bisectors  of  the  sides 
of  a  triangle.  Do  they  meet  in  a  point  ? 

Ex.  8.  Draw  accurately  the  three  bisectors  of  the  angles  of  a  triangle. 
Do  they  meet  in  a  point  ? 


CONSTRUCTION   PROBLEMS 


123 


Draw  line  AB  _L  to 

Q.E.F. 

(Const.). 
(202).      Q.E.D. 


PROPOSITION  XXXVIII.     PROBLEM 

262.  To  draw  a  tangent  to  a  given  circle  through  a  given 
point :  A- 

I.    If  the  point  is  on  the  circle. 
II.    If  the  point  is  without  the  circle. 

I.  Given:    O  O;  P,  a  point  on  the 
circle. 

Required:  To  draw  a  tangent 
through  P. 

Construction:  Draw  the  radius  OP. 
OP  at  P  (by  257). 

Statement :  AB  is  tangent  to  0  O  at  P. 

Proof :  AB  is  _L  to  OP  at  P 

.-.  AB  is  a  tangent 

II.  Given:  O  O;  P,  a  point  with- 
out it. 

Required:  To  draw  a  tangent 
through  P. 

Construction:  DrawPO;  bisect  it  at 
Jf(by  254). 

Using  M  as  a  center  and  PM  as  a  radius,  describe  a  circum- 
ference intersecting  O  O  at  A  and  B. 

Draw  PA,  PB,  OA,  OB. 

Statement:   PA  and  PB  are  tangents  through  P         Q.E.F. 

Proof :  O  M  passes  through  O  (PJf  =  MO  by  const.). 

.•.  <  P^IO  is  a  rt.  Z  (240). 

.-.  PA  is  a  tangent  (202). 

Similarly,  PB  is  a  tangent.  •  Q.E.D. 

NOTE.  The  student  has  probably  observed  that  in  constructions  cer- 
tain lines  and  angles  must  precede  others.  The  order  of  the  successive 
steps  is  an  important  consideration.  Thus,  in  261  and  262  certain  lines 
must  be  drawn  before  others  can  be  drawn. 


124 


BOOK   II.    PLANE  GEOMETRY 


PROPOSITION  XXXIX.    PROBLEM 

263.  To  circumscribe  a  circle  about  a  given  triangle. 
Given:   (?). 

Required:   (?).     (See  214.) 

Construction:  Bisect  AB,  BC,  AC. 
Erect  Js  at  T,  #,  s,  meeting  at  o. 

Using  o  as  a  center  and  OA  as  radius, 
draw  a  circle. 

Statement :  This  O  will  pass  through 
vertices  A,  .B,  and  c.  Q.E.F. 

Proof:   (100). 

PROPOSITION   XL.     PROBLEM 

264.  To  inscribe  a  circle  in  a  given  triangle. 


Q.E.D, 


M 

Given:  (?).   Required:  (?). 

Construction:  Draw  the  three  bisectors  of  the  A  of  A  ABC, 
meeting  at  o  (by  256).  Draw  Js  from  O  to  the  three  sides. 

Using  O  as  a  center  and  one  -L  as  a  radius,  draw  a  O. 

Statement:  This  O  will  be  tangent  to  the  three  sides  of 
A  ABC.  Q.E.F. 

Proof:   The  bisectors  of  these  angles  meet  in  a  point  and 

the  Js  OX,  OM,  ON  are  equal  (99). 

.*.  the  circumference  passes  through  i,  jf,  N    (1T9). 

Also  the  three  sides  are  tangent  to  the  O          (202). 

That  is,  the  O  O  fe  inscribed  in  A  ABC  (221). 

Q.E.D. 


CONSTRUCTION   PROBLEMS  125 


* 


265.  If  a  circle  is  described  tangent  to  one  side  of  a  triangle 
and  tangent  to  the  prolongations  of  the  other  sides,  it  is  called 
an  escribed  circle.     Every  triangle  may  have  three  escribed 
circles. 

PROPOSITION  XLI.     PROBLEM 

266.  To  construct  a  parallelogram  if  two  sides  and  the 
included  angle  are  given.  w  z 

Given:    The    sides    a   and   b   and  /  / 

their  included  angle,  x. 

Required:   To  construct  a  O  con- 
taining these  parts. 

Construction:    Take  a  straight  line 
PQ  =  to  a. 

At  P  construct  Z.  p  =  to  Z  x.     On  PIT,  the  side  of  this  Z, 
take  PR  =  to  b. 

At  R  draw  RY  II  to  PQ ;  and  at  Q  draw  QZ  II  to  PW. 

Denote  the  intersection  of  these  lines  by  8. 

Statement:   PQSR  is  the  required  parallelogram.        Q.E.F. 

Proof:    First,  it  is  a  parallelogram.  (Def.). 

Second,  it  is  the  required  parallelogram.     (Because  it  con- 
tains the  given  parts.)  Q.E.D. 


tji 

p/        a  I     /*• 


Ex.  1.  Draw  a  triangle  and  all  its  exterior  angles.  Bisect  these  to  find 
centers  of  escribed  circles.  What  are  the  radii  of  these  circles?  Draw 
the  three  escribed  circles. 

Ex.  2.    Draw  a  rectangle,  having  given  two  sides. 

Ex.  3.  What  several  things  must  be  known  about  a  circle  before  you. 
can  draw  a  tangent  ? 

Ex.  4.    Is  a  radius  of  a  circle  a  tangent  ?     Why? 

Ex.  6.  How  can  one  find  the  center  of  a  given  arc  ?  Is  this  the  same 
as  its  midpoint  ? 

Ex.  6.   Give  another  method  of  solving  Proposition  XXXVII. 

Ex.  7.  If  a  hundred  triangles  stood  on  the  same  base,  and  all  their 
vertex  angles  were  equal,  where  would  all  their  vertices  be  ? 


126  BOOK   II.    PLANE   GEOMETRY 

PROPOSITION  XLII.     PROBLEM 

267.    To  construct  a  segment  of  a  circle  upon  a  given  line,  as 
chord,  which  shall  contain  angles  equal  to  a  given  angle. 
Given:    Line  AB  and  Z  K' . 
Required:    To  construct   a  seg- 
ment upon  AB  the  inscribed  angles 
of  which  shall  =  Z  Kf. 

Construction:    Construct    at   A, 
Z  BAC=  to  Z.K'. 

Bisect  AB  at  M. 
At  M  erect  OM  _L  to  AB. 
At  A  erect  OA  _L  to  AC,  meet- 
ing OM  at  O. 

Using  O  as  a  center  and  OA  as  radius,  describe  O  O. 
Statement :  The  A  inscribed  in  segment  AKB  =  Z  K1.  Q.E.F. 
Proof:    The  circle  passes  through  B  (80). 

.-.  AB  is  a  chord  (181). 

AC  is  a  tangent  to  the  O  (202). 

.*.  Z  BAG  is  meas.  by  half  the  arc  AB  (241). 

Any  angle  inscribed  in  AKB  is  meas.  by  half  arc  AB 

(236). 

.-.  any  angle  AKB  =  Z  BAG  (237). 

.-.  any  inscribed  Z  AKB  =  Z  K'  (Ax.  1). 

Q.E.D. 

[Let  the  pupil  draw  chords  A  K  and  BK,  which  were  purposely 
omitted.] 

Historical  Note.  Substantial  contributions  were  made  to  the  ad- 
vancement of  geometrical  science  by  Hippocrates,  a  Greek  philosopher, 
who,  during  the  fifth  century  B.C.,  discovered  many  properties  of  the  circle. 
He  was  the  first  to  employ  the  method  of  proof  known  as  the  reductio  ad 
absurdum,  and  he  wrote  the  first  text  book  on  geometry.  He  was  not 
aware  of  the  truth  that  equal  central  angles  and  equal  inscribed  angles 
intercept  equal  arcs,  although  he  knew  that  areas  of  circles  are  propor- 
tional to  the  squares  of  their  radii. 


CONSTRUCTION  PROBLEMS  127 

PROPOSITION  XLIII.     PROBLEM 

268.  To  construct  the  third  angle  of  a  triangle  if  two  angles 
are  known. 

Given  :    z§  A  and  B,  two  A  of  a  A. 
Required  :    To  construct  the  third. 
Construction:   At  point  O  in  line  RS 
construct  Z  a  =  to  Z  A. 

At  point  O  in  OT  construct  Z  b  =  to  Z  B. 
Statement:   The  Z  VOR  =  the  third  Z  of  the  A.         Q.E.F. 
Proof  :  Z  a  -f  Z  6  +  Z  FOR  =  2  rt.  Z  (46). 

Z^i  +  Z  B  -f  the  third  Z  of  the  A  =  2  rt.  ^         (104). 
.-.  Z  «  -f-  Z  6  -|-  Z  FOR  =  Z  ^4  4-  Z  J5  4-  the  third  Z 

(Ax.  1). 

But  Za  +  Z£  =Z^1  +  Z5  (Const,  and  Ax.  2). 

Subtracting,          Z  FOR=  the  third  Z  of  the  A     (Ax.  2). 

Q.E.D. 

PROPOSITION  XLIV.     PROBLEM 

269.  To  construct  a  triangle  if  the  three  sides  are  known. 
Given  :  Sides  a,  5,  c  of  a  A.  T 
Required  :    To    construct     ?^ 

the  A.  c 

Construction  : 
Draw       RS  =  to  a.  4L  _  2. 

R  Q 

Using  R  as  a  center  and  b 

as  a  radius,  describe  an  arc.  Using  5  as  a  center  and  c  as  a 
radius,  describe  an  arc  intersecting  the  former  arc  at  T. 
Draw  RT  and  ST. 


Statement:         A  #sr  is  the  required  A.  Q.E.F. 

Proof:  RSTisaA  (23). 

.R/ST  is  the  required  A.     (It  contains  a,  6,  <?.)     Q.E.D. 
Discussion  :   Is  this  problem  ever  impossible  ?     When  ?  . 


128 


BOOK   II.    PLANE   GEOMETRY 


Ex.  1.    Construct  an  angle  of  60° ;  of  30° ;  of  15° ;  of  7°  30' ;  of  75°. 

Ex.  2.   Trisect  a  right  angle. 

Ex.  3.    Construct  a  tangent  to  a  circle,  parallel  to  a  given  line. 

Ex.  4.    Construct  a  tangent  to   a  given   circle   perpendicular   to    a 
given  line. 

Ex.  6.    Construct  through  a  given  point  within   a 
circle,  two  chords  each  equal  to  a  given  chord.     Is  this     A  i 
ever  impossible  ? 

Ex.  6.    Construct  in  a  given  circle  a  chord  equal  to 
a  second  chord  and  parallel  to  a  third. 

PROPOSITION  XLV.     PROBLEM 

270.  To  construct  a  triangle  if  two  sides  and  the  included 
angle  are  known. 

Given :   The  sides  a  and  6,  and  their 
included  Z  c  in  a  A. 

Required :   To  construct  the  A. 

Construction:  Draw  CB  =  to  a.    Ate     c/          ° D 

construct  Z  BCX  =  to  given  Z  c.     On 
CX  take  CA  =  to  b.     Join  AB. 

Statement:   (?).     Proof:   (?).     Discussion:   (?). 

PROPOSITION  XLVI.     PROBLEM 

271.  To  construct  a  trjangle  if  a  side  and  the  two  angles  ad- 
joining it  are  known. 

Y\AxX 


/ 

c^  _ 


Given:  (?).     Required:  (?). 

Construction :  Draw  BC  =  to  a.  At  B  construct  Z  CBX  = 
to  Z  Bf  ;  at  C  construct  Z  BCY  =  to  Z  cf.  Denote  the  point 
of  intersection  of  BX  and  CY  by  A. 

'Statement:  (?).     Proof:  (?).    Discussion:  (?). 


CONSTRUCTION   PROBLEMS  129 


PROPOSITION  XLVII.     PROBLEM 

272.    To  construct  a  right  triangle  if  the  hypotenuse  and  a 
leg  are  known.  c 


Given :  Hypotenuse  c  ;  leg  b. 

Required:  (?). 

Construction:    Draw  an   indefinite     c 


line  CD  and  at  C  erect  a  J_  =  to  b. 

Using  A  as  a  center  and  c  as  a  radius,  describe  an  arc  cutting 

CD  at  B.     Draw  AB. 

Statement:  (?).     Proof:  (?).     Discussion:  (?). 

Another  Construction :    Take  a  line,  ES  =  to  c.     With  its 
midpoint  Jf,  as  center,  and  EM  as  radius 
describe  a  semicircle.     With  E  as  center 
and  b  as  radius,  describe  an  arc  cutting 
the  semicircumference  at  T.      Draw  TE 

ITI 

and  TS. 

Statement:  (?).     Proof:  (?).     Discussion:  (?). 

PROPOSITION  XLVIII.     PROBLEM* 

273.  To  construct  a  triangle  if  an  angle,  a  side  adjoining  it, 
and  the  side  opposite  it  are  known  ;  that  is,  if  two  sides  and 
an  angle  opposite  one  of  them  are  known. 

The  known  angle  may  be  obtuse,  right,  or  acute.    Consider  : 
First,  If  "side  opposite"  >  "side  adjoining." 
Second,  If  "side  opposite"  =  "  side  adjoining." 
Third,  If  "  side  opposite  "  <  "  side  adjoining." 
Construction  for  all :  Draw  an  indefinite  line,  CX,  and  at 
one  extremity  construct  an  Z  =  to  Z.  C.     Take  on  the  side 
of  this  angle  a  distance  from  the  vertex  equal  to  the  "  side 
adjoining."     Using  the  end  of  this  side  as  a  center  and  the 
"  side  opposite  "  as  a  radius,  describe  an  arc  intersecting  CX. 
Draw  radius  to  the  intersection  just  found. 


130 


BOOK   II.     PLANE   GEOMETRY 


If  the  known  angle  is  obtuse  or  right. 

Given :   Z  c,  side  c  opposite    y 
it,  and  side  b  adjoining  Z  C. 

Construction:  As  above. 

Discussion :   Case  I.     c  >  b. 
The  A  is  always  possible. 
»  Case  II.     c  =  b. 

The  A  is  never  possible  (106). 

Case  III.     c<b. 

The  A  isjiever  possible  (116). 

If    the    known    angle    is 
acute. 

Case  I.     c>  b. 

The  A  is  always  possible. 

Case  II.     c  =  b. 

The  A  is  always  possible 
and  isosceles. 

Case  III.     c  <  b. 

First,  c  <.the  J_  from  A  to  CX. 

The  A  is  never  possible. 


Second,  c  —  the  -L  from  A  to  CX. 
The  A  is  possible  and  a  right  A. 


Third,  c  >  the  _L  from  A  to 
CX. 

There  are  two  A,  ACB  and 
ACE'  ;  both  contain  the  three 
given  parts. 


,Y 


ANALYSIS 


131 


ANALYSIS 

Many  constructions  are  so  simple  that  their  correct  solu- 
tion will  readily  occur  to  the  pupil. 

Sometimes,  in  the  case  of  complicated  constructions,  the 
pupil  must  have  the  ability  to  put  the  given  parts  together, 
one  by  one.  The  following  outline  may  be  found  helpful  if 
employed  intelligently.  It  is  called  the  method  of  analysis. 

I.  Suppose  the  construction  made,  that   is,  suppose  the 
figure  drawn. 

II.  Study  this  figure  in  search  of   truths  by  which  the 
order  of  the  lines  that  have  been  drawn  can  be  determined. 

III.  One  or  more  auxiliary  lines  may  be  necessary. 

IV.  Finally,  construct  the  figure  and  prove  it  correct. 

EXERCISE.  Given  the  base  of  a  triangle,  an  adjacent  acute 
angle,  and  the  difference  of  the  other  sides,  to  construct  the 
triangle. 

Given :  Base  AB  ;  Z  A' ;  difference  d. 

Required :  To  construct  the  A. 

[Analysis:  Suppose  A  ABC  is  the  re- 
quired A.  It  is  evident  if  CD  =  CB,  they 
may  be  sides  of  an  isos.  A  and  AD  =  d.~\ 

Construction :  At  A  on  AB  construct 
Z  BAX  =  toZAr  and  on  AX,  take  AD  = 
to  d.  Join  DB.  At  M,  midpoint  of 
DB,  draw  M Y  _L  to  DB  meeting  AX  at 
Draw  CB. 


C. 


Statement: 


Proof:  (?).    Discussion: 


Historical  Note.  The  philosopher  Plato  and  his  school  flourished  at 
Athens  in  the  fourth  century  B.C.  Plato  brought  to  geometry  exact 
definitions  and  axioms  as  well  as  the  method  of  analysis,  which  is  helpful 
in  discovering  difficult  proofs  and  constructions. 


132 


BOOK   II.    PLANE   GEOMETRY 


ORIGINAL  CONSTRUCTIONS 

I.    Construct  an  isosceles  triangle,  having  given: 

1.  The  base  and  one  of  the  equal  sides. 

2.  The  base  and  one  of  the  equal  angles. 

3.  One  of  the  equal  sides  and  the  vertex  angle. 

4.  One  of  the  equal  sides  and  one  of  the  equal  angles. 

5.  The  base  and  the  altitude  upon  it. 

6.  The  base  and  the  radius  of  the  inscribed  circle. 
[Bisect  the  base  ;  erect  a  JL  =  to  the  radius  ;  describe  O,  etc.] 

7.  The  base  and  the  radius  of  the  circumscribed  circle. 

8.  The  altitude  and  the  vertex  angle. 

9.  The  base  and  the  vertex  angle. 
[Find  the  supplement  of  the  given 

base  construct  an  Z  =  to  this  half  ;  etc.] 

10.   The  perimeter  and  the  altitude. 

Given  :  Perimeter  =  AB  ;  alt.  =  h.  Re- 
quired: (?).  Construction:  Bisect  AB; 
erect  at  M  _L  =  to  h  ;  draw  «4P  and  BP. 
Bisect  these  ;  erect  Js  SC  and  RE  ;  etc. 


bisect  this;  at  each  end  of 


C    M     E 


II.    Construct  a  right  triangle,  having  given: 

11.  The  two  legs. 

12.  One  leg  and  the  adjoining  acute  angle. 

13.  One  leg  and  the  opposite  acute  angle. 
The  hypotenuse  and  an  acute  angle. 


14. 


16.    The  hypotenuse  and  the  altitude  upon  it. 

16.    The  median  and  the  altitude  upon  the  hypotenuse 


17.  The  radius  of  the  circumscribed  circle  and  a  leg. 

18.  The  radius  of  the  inscribed  circle  and  a  leg. 
Given:   Radius  =  r;   leg  =  CA.     Required:    (?). 

Analysis :  Consider  that  ABC  is  the  completed  fig- 
ure ;  CNOM  is  a  square,  whose  vertex  O  is  the  center 
of  the  circle,  and  side  ON  is  the  given  radius.  AB  is 
tangent  from  A .  Construction:  On  CA  take  CN  = 
to  r  and  construct  square,  CNOM.  Prolong  CM  in- 
definitely. Describe  O,  etc. 


ORIGINAL  CONSTRUCTIONS 


133 


B  .„- 


19.  One  leg  and  the  altitude  upon  the  hypotenuse. 

20.  An  acute  angle  and  the  sum  of  the  legs. 
Given:    AD  =  sum;    Z  K.      Required:      (?). 

Construction  :  At  A  construct  /.  A  =  to  Z  K ;  at  D 

construct  Z  D  =  to  45°,  etc.  

^  C 

21.  The  hypotenuse  and  the  sum  of  the  legs. 

[Use  A  as  center,  hypotenuse  as  radius,  etc.] 

22.  The  radius  of  the  circumscribed  circle  and  an  acute  angle. 

23.  The  radius  of  the  inscribed  circle  and  an  acute  angle. 
Construction :  Take  CS  =  to  r,  on  indefinite 

line,  ZA.  On  CS  construct  square  CSOM.  At 
O  construct  Z  MOX  =  to  Z  K.  Draw  radius  OT 
_L  to  OX.  Draw  tangent  at  T,  etc. 


III.  Construct  an  equilateral  triangle,  having  given  : 

24.  One  side. 

25.  The  altitude. 

26.  The  perimeter. 

27.  A  median. 

28.  The  radius  of  the  inscribed  circle. 
[Draw  circle  and  radius ;  at  center  construct 

=  to  120°  and  Z  ROT  =  to  120° ;  etc.] 

29.  The  radius  of  the  circumscribed  circle. 


IV.    Construct  a  triangle,  having  given : 

30.  The  base,  an  angle  adjoining  it,  the  altitude  upon  it. 

31.  The  midpoints  of  the  three  sides. 
[Draw  RS,  RT,  ST,  etc.] 

32.  One  side,  altitude  upon  it,  and  the  radius 
of  the  circumscribed  circle. 

Construction:   Draw  O  with  given  radius  and 
any  center.     Take  chord  =  to  given  side ;  etc. 

33.  One  side,  an  adjoining  angle,  and  the  radius  of  the  circumscribed 
circle. 

34.  Two  sides  and  the  altitude  from  the  same  vertex. 
Construction:  Erect  JL  equal  to  altitude,  upon  an  indefinite  line.     Use 

the  end  of  this  altitude  as  center,  and  the  given  sides  as  radii,  etc. 


134 


BOOK   II.     PLANE   GEOMETRY 


<v 

R 


>C 


/     KX::=:V*C 

L.L  yt\ 


lih 

:  • 

\i^__ 

Av 


c.. 


36.  One  side,  an  angle  adjoining  it,  and  the 
sum  of  the  other  two  sides. 

Construction  :  At  A  construct  /.BAX  =  to  given 
ZK.  On  AX  take  AD  =  to  s;  draw  DB  ;  bisect 
DB  at  M,  etc. 

36.  Two  sides  and  the  median  to  the  third  side. 
Given :  a,  b,  m .    Construction :  Con  struct  A  A  BR 

with  three  sides,  A  B  =  to  a,  BR  =  to  b,  AR  =  to 
2  m.  Draw  A  C  II  to  BR  and  .RC  II  to  ,4£  meeting 
atC.  Draw^C.  Statement:  (?).  Proof:  (?). 

37.  A  side,  the  altitude  upon  it,  and  the  angle 
opposite  it. 

Given :   Side  =  to  AB,   alt.  =  to  h  ;  opposite 
Z  C  =  to  Z  C". 

Construction:   Upon  AB  construct   segment 
ACB  which  contains  Z  C  =  to  Z  C'  (by  267). 
At  A  erect  ^4/2  _L  to  ^£  and  =  to  h  ;  etc. 

38.  A  side,  the  median  to  it,  the  angle  oppo- 
site it. 

[Statement :  A  ABC  is  the  required  A.] 

39.  One  side  and  the  altitude  from  its  extrem- 
ities to  the  other  sides. 

Given:  Side  =  AB,  altitudes  x  and  y. 

Construction:  Bisect  AB]  describe  a  semicircle. 
Using  A  as  center  and  x  as  radius,  describe  arc  cutting 
the  semicircle  at  R  ;  etc. 

40.  Two  sides  and  the  altitude  upon  one  of  them. 
[Given :  Sides  =  to  AB  and  EC ;  alt.  on  EC  =  to  x.~\ 

41.  One  side,  an  angle  adjoining  it,  and  the  radius  of  the  inscribed  circle. 
Construction :  Describe  O  with  given  radius,  any  center. 
Construct  central  Z  =  to  given  Z.     Draw   two  tangents  ||  to  these 

radii;  etc.  

V.  Construct  a  square,  having  given : 

42.  One  side. 

43.  The  diagonal. 

44.  The  perimeter. 

45.  The  sum  of  a  diagonal  and  a  side. 


ORIGINAL  CONSTRUCTIONS  135 

VI.    Construct  a  rhombus,  having  given : 

46.  One  side  and  an  angle  adjoining  it. 

47.  One  side  and  the  altitude. 

48.  The  diagonals. 

49.  One  side  and  one  diagonal.     [Use  269.] 

60.   An  angle  and  the  diagonal  to  the  same  vertex. 

51.  An  angle  and  the  diagonal  between  two  other  vertices 

52.  One  side  and  the  radius  of  the  inscribed  circle. 


VII.    Construct  a  rectangle,  having  given  : 

53.  Two  adjoining  sides. 

54.  A  diagonal  and  a  side. 

55.  One  side  and  the  angle  formed  by  the  diagonals. 

56.  A  diagonal  and  the  sum  of  two  adjoining  sides.     [See  Ex.  21.] 
67.  A  diagonal  and  the  perimeter. 

58.    The  perimeter  and  the  angle  formed  by  the  diagonals. 

Construction  :  Bisect  the  perimeter  and  V.. 

take  AB  =  to    half  of    it.      Bisect  Z  K.  ^—  -  —  ^K 

At  A   construct   ZBAX  =  to    half    /.K.  " 

Etc. 


VIII.    Construct  a  parallelogram,  having  given  : 

69.  One  side  and  the  diagonals. 

60.  The  diagonals  and  the  angle  between  them. 

61.  One  side,  an  angle,  and  the  diagonal  not  to  the  same  vertex. 

62.  One  side,  an  angle,  and  the  diagonal  to  the  same  vertex. 

63.  One  side,  an  angle,  and  the  altitude  upon  that  side. 

64.  Two  adjoining  sides  and  the  altitude. 


IX.    Construct  an  isosceles  trapezoid,  having  given : 

65.  The  bases  and  an  angle  adjoining  the  larger  base. 

66.  The  bases  and  an  angle  adjoining  the  less  base. 

67.  The  bases  and  the  diagonal. 

68.  The  bases  and  the  altitude. 


136  BOOK   II.     PLANE   GEOMETRY 

69.  The  bases  and  one  of  the  equal  sides. 

70.  One  base,  an  angle  adjoining  it,  and  one  of  the  equal  sides. 

71.  One  base,  the  altitude,  and  one  of  the  equal  sides. 

72.  One  base,  the  radius  of  the  circumscribed  circle,  and  one  of  the 
equal  sides.     [First,  describe  a  O.] 

73.  One  base,  an  angle  adjoining  it,  and  the  radius  of  the  circum- 
scribed circle. 

74.  The  bases  and  the  radius  of  the  circumscribed  circle. 

75.  One  base  and  the  radius  of  the  inscribed  circle. 
Construction :  Bisect  the  base  and  erect  a  _L  =  to  radius ;  etc. 


X.  Construct  a  trapezoid,*  having  given  : 

76.  The  bases  and  the  angles  adjoining  one  of  them. 
Construction :   Take  EC  =  to  longer  base,  and  on  it  take  ED  =  to 

less  base.     Construct  A  DBC  (by  271). 

77.  The  four  sides. 

78.  A  base,  the  altitude,  and  the  non-parallel  sides. 
Construction:  Construct  a  A  two  sides  of  which  equal  the  given  non-|| 

sides  of  the  trapezoid,  and  the  altitude  from  same  vertex  of  which  equals 
the  given  altitude.     (See  Ex.  34.) 

79.  The  bases,  an  angle,  and  the  altitude. 

Construction:  Construct  £7  on  ED,  having  given  altitude  and  £. 

80.  A  base,  the  angles  adjoining  it,  and  the  altitude. 

81.  The  longer  base,  an  angle  adjoining  it,  and  the  non-parallel  sides. 

82.  The  shorter  base,  an  angle  not  adjoining  ;t,  and  the  non-parallel 
sides. 

XI.  Construct  the  locus  of  a  point  which  shall  be : 

83.  At  a  given  distance  from  a  given  point. 

84.  At  a  given  distance  from  a  given  line. 

*  NOTE.     It  is  evident  that  every  trapezoid  may       ^ B 

be  divided  into  a  parallelogram  and  a  triangle  by 
drawing  one  line  (as  BD]  II  to  one  of  the  non-11  sides 
Hence  the  construction  of  a  trapezoid  is  often  merely 
constructing  a  triangle  and  a  parallelogram. 


:r/\ 


ORIGINAL  CONSTRUCTIONS  137 

86.    At  a  given  distance  from  a  given  circle : 

(£)    If  the  given  radius  is  <  the  given  distance ; 
(ii)  If  the  given  radius  is  >  the  given  distance. 

86.  Equally  distant  from  two  given  points. 

87.  Equally  distant  from  two  intersecting  lines. 


XII.    Find    (by  intersecting    loci)  *  the  point   P,  which 
shall  be : 

88-   At  two  given  distances  from  two  given  points. f 

89.  Equally  distant  from  three  given  points. 

90.  In  a  given  line  and  equally  distant  from  two  given  points. 

91.  In  a  given  line  and  equally  distant  from  two  given  intersecting 
lines. 

92.  In  a  given  circle  and  equally  distant  from  two  given  points.f 

93.  In    a  given   circle   and   equally  distant  from  two  intersecting 
lines,  f 

94.  Equally  distant  from  two  given  intersecting  lines  and  equally 
distant  from  two  given  points.f 

95.  At  a  given  distance  from  a  given  line  and  equally  distant  from 
two  given  points.f 

96.  At  a  given  distance  from  a  given  line  and  equally  distant  from 
two  other  intersecting  lines. f 

97.  Equally  distant  from  two  given  points  and  at  a  given  distance 
from  one  of  them.f 

98.  Equally  distant  from  two  given  intersecting  lines  and  at  a  given 
distance  from  one  of  them.f 

99.  At  a  given  distance  from  a  point  and  equally  distant  from  two 
other  points. f 

100.  At  given  distances  from  two  given  intersecting  lines. f 

101.  At  given  distances  from  a  given  line  and  from  a  given  circle.f 

*  It  is  well  to  draw  the  loci  concerned  as  dotted  lines.     (See  Ex.  105.) 
t  In  the  Discussion,  include  the  answers  to  questions  like  these  : 

(1)  Is  this  ever  impossible  ?  (i.e.  must  there  always  be  such  a  point  ?) 

(2)  Are  there  ever  two  such  points  ?  when  ? 

(3)  Are  there  ever  more  than  two  ?  when  ? 

(4)  Is  there  ever  only  one  ?  when  ? 

ROBBINS'S    NEW   PLANE    GEOM.  —  10 


138  BOOK   II.     PLANE   GEOMETRY 

102.  At  given  distances  from  a  given  line  and  from  a  given  point.* 

103.  Equally  distant  from  two  parallels  and  equally  distant  from 
two  intersecting  lines.* 

104.  At  a  given  distance  from  a  given  point  and  equally  distant  from 
two  given  parallels.*  , 

105.  At  a  given  distance  from  a  given  point 

and  equally  distant  from  two  given  intersecting  p^v^x^  "\  p 
lines. 

Can  C  be  so  taken  that  there  will  be  no  point? 

Can  C  be  so  taken  that  there  will  be  only  one  \              \/ 

point?  only  two?  only  three?  more  than  four?  ''-•:- •• 


XIII.    Find  (by  intersecting  loci)  the  center  of  a  circle 
which  shall : 

106.  Pass  through  three  given  points. f 

107.  Pass  through  a  given  point  and  touch  a  given  line  at  a  given 
point,  f 

108.  Have  a  given  radius  and  be  tangent  to  a  given  line  at  a  given 
point.f 

109.  Have  a  given  radius,  touch  a  given  line,  and  pass  through  a  given 
point.f 

110.  Pass  through  a  given  point  and  touch  two  given  parallel  lines. f 

111.  Touch  two  given  parallels,  one  of  them  at  a  given  point,  f 

112.  Have  a  given  radius  and  touch  two  given  intersecting  lines,  t 

113.  Have  a  given  radius  and  pass  through  two  given  points. f 

114.  Touch  three  given  indefinite  lines,  no  two  of  them  being  parallel. :{: 
116.   Touch  three  given  lines,  only  two  of  them  being  parallel. 


XIV.    Construct  a  circle  which  shall : 

116.  Pass  through  a  given  point  and  touch  a  given  line  at  a  given  point. 

117.  Touch  two  given  parallel  lines,  one  of  them  at  a  given  point. 

118.  Pass  through  a  given  point  and  touch  two  given  parallels. 

*  See  note  (t)  on  preceding  page. 

t  Discussion :  Is  this  ever  impossible  ?    Are  there  ever  two  circles  and  hence 
two  centers  ?     Are  there  ever  more  than  two  ?     Etc. 
\  Four  solutions.     One  is  in  265. 


ORIGINAL  CONSTRUCTIONS 


139 


119.  Have  a  given  radius,  touch  a  given  line,  and  pass  through  a 
given  point. 

120.  Have  its  center  in  one  line,  touch  another  line,  and  have  a  given 
radius. 

121.  Have  a  given  radius  and  touch  two  given  intersecting  lines. 

122.  Have  a  given  radius  and  pass  through  two  given  points. 

123.  Have   a  given   radius    and   touch   a  given   circle    at    a    given 
point.     [Draw  tangent  to  the  given  O  at  the  given  point.] 

124.  Have  a  given  radius  and  touch  two  given  circles. 

125.  Touch  three  indefinite  intersecting  lines.* 

126.  Touch  two  given  intersecting  lines,  one  of  them  at  a  given 
point. 

127.  Touch  a  given  line  and  a  given  circle 
at  a  given  point. 

Given:  Line  AB;  O  C;  point  P. 

Construction :  Draw  radius  CP.  Draw  tan- 
gent at  P  meeting  AB  at  R.  Bisect  /.PRB, 
meeting  CP  produced  at  0;  etc. 

128.  Be  inscribed  in  a  given  sector. 

Construction :  Produce  the  radii  to  meet  the  tangent  at  the  midpoint 
of  the  arc.     In  this  A  inscribe  a  O. 

129.  Have  a  given  radius  and  touch  two  given  circles. 

130.  Have  a  given  radius,  touch  a  given  line,  and  a  given  circle, 

131.  Touch  a  given  line  at  a  given  point 
and  touch  a  given  circle. 

Given:  Line  AB]  point  P;  O  C. 
Construction :  At  P  erect  PX  ±  to  AB,  and 
extend  it  below  AB,  so  PR  =  radius  O  C. 
Draw  CR  and  bisect  it  at  M. 


Erect  MY  ±  to  CR  at  M,  meeting  PX  at  0 ; 


etc. 


Y 


132.  What  is  the  locus  of  the  vertices  of  all  right  triangles  having 
the  same  hypotenuse? 

133.  Through   a  given   point  on  a  given   circumference   draw  two 
equal  chords  perpendicular  to  each  other. 


*  Four  solutions.     One  is  in  265. 


140 


BOOK   II.     PLANE   GEOMETRY 


134.  Draw  a  line  of  given  length  through  a  given  point  and  terminat- 
ing in  two  given  parallels. 

Construction:  Use  any  point  of  one  of  the  lls  as  center  and  the  given 
length  as  radius  to  describe  an  arc  meeting  the  other  II.  Join  these  two 
points.  Through  the  given  point  draw  a  line  II,  etc. 

135.  Draw  a  line,  terminating  in  the  sides     Q 

of  an  angle,  which  shall  be  equal  to  one  line       x/ 

and  parallel  to  another.  /    ./^      / 

Statement :  RS  =  a,  and  is  II  to  x. 

136.  Draw  a  line  through  a  given  point  within  an  angle,  which  is 
terminated  by  the  sides  of  the  angle  and  bisected  by  the  point. 

Construction :  Through  P  draw  PD  II  to 
AC.  Take  on  AB,  DE  =  AD.  Draw 
EPF;  etc. 

137.  Circumscribe  a  circle  about  a  rec- 

tangle. 

138.  Construct  three  circles  having  the  vertices  of  a  given  triangle  as 
centers  so  that  each  touches  the  other  two. 

Construction :  Inscribe  a  O  in  the  A ;  etc. 

139.  Construct  within  a  circle  three  equal  circles 
each  of  which  touches  the  given  circle  and  the  other     A 
two. 

Construction:   Draw  a  radius,   OA,  and  construct 
Z  A  OB  =  to  120°  and  Z.  A  OC  =  to  120°.     In  these  sectors  inscribe,  etc. 

140.  Through  a  point  without  a  circle  draw  a 
secant  having  a  given  distance  from  the  center. 

141.  Draw  a  diameter  to  a  circle  at  a  given  dis- 
tance from  a  given  point. 

142.  Through  two  given  points  within  a  circle 
draw  two  equal  and  parallel  chords. 

Construction:    Bisect   the    line   joining   the 
given  points  and  draw  a  diameter,  etc. 

143.  Draw  a  parallel  to  side  EC  of  triangle 
ABC,  meeting  AB  in  X  and  AC  in  F,  such  that 
XY  =  YC.  l 


ORIGINAL  CONSTRUCTIONS 


141 


144.   Find  the  locus  of  the  points  of  contact  of  the  tangents  drawn  to  a 
series  of  concentric  circles  from  an  external  point. 

146.    Given :  Line  AB  and  points  C  and  D  on 
the  same  side  of  it ;  find  point  X  in  AB  such 


Construction:   Draw   CE  _L  to  AB  and  pro- 
duce to  F  so  that  EF  =  CE.    Draw  FD  meeting 
AB  in  X.     Draw  CX. 

146.  Draw  from  one  given  point  to  another  the 
shortest  path  which  has  one  point  in  common  with 
a  given  line. 

Statement :  CX  +  XD  is  <  CR  +  RD. 

Another  statement  of  this  exercise:  If  C  is  an 
object  before  a  mirror  ER,  and  D  is  an  eye,  draw  a 
diagram  showing  the  path  of  a  ray  of  light,  from  C,  reflected  to  D. 

147.  Draw  a  line  parallel  to  side  BC  of  triangle  ABC  meeting  AB  at 
X  and  A  C  at  F,  so  that  XY  =  BX  +  YC. 

Construction :  Draw  bisectors  of  A  B  and  C,  meeting  at  0,  etc. 

148.  Draw  in  a  circle,  through  a  given  point  of  an  arc,  a  chord  that 
is  bisected  by  the  chord  of  the  arc. 

Construction :  Draw  radius  OP  meeting  chord  at  C. 
Prolong  PO  to  X  so  that  CX  =  OP.  Draw  XM  II  to 
AB  meeting  O  at  M.  Draw  PM  cutting  AB  at  Z>; 
etc.  Is  there  any  other  chord  from  P  bisected  by  AB  ? 

149.  Inscribe  in  a  given  circle  a  triangle  whose 
angles  are  given. 

Construction :  Construct  3  central  A,  doubles  of  the  given  A.    Etc. 

160.    Circumscribe  about  a  given  circle  a  triangle  whose  angles  are 
given. 

Construction :  Inscribe  A  (like  Ex.  149)  first,  and  draw  tangents  II  to 
the  sides. 

151.  Three  lines  meet  in  a  point.  Draw  a  line 
terminating  in  the  outer  two  and  bisected  by  the 
inner  one. 

Construction :  Through  any  point  P,  of  OB,  draw 
Us  to  the  outer  lines.  Draw  diagonal  RS ;  etc. 


142 


BOOK   II.     PLANE   GEOMETRY 


152.  Draw  through  a  given  point  P,  a  line  that 
is  terminated  by  a  given  circle  and  a  given  line 
and  is  bisected  by  P. 

Construction:  Draw  any  line  DX  meeting  AB 
at  D.  Draw  PE  II  to  AB  meeting  DX  at  E.  Take 
EF  =  ED  ;  etc. 

153.  Through  a  given  point  without  a  circle  draw  a  secant  to  the 
circle  which  is  bisected  by  the  circle. 

Construction:  Draw  arc  at  J",  using  P  as  center 
and  diam.  of  O  O  as  radius.  Using  T  as  center 
and  same  radius  as  before,  describe  circle  touching 
O  0  at  C  and  passing  through  P.  Draw  PC  meet- 
ing O  0  at  M. 

154.  Inscribe  a  square  in  a  given  rhombus. 
[Bisect  the  four  A  formed  by  the  diagonals.] 

155.  Bisect  the  angle  formed  by  two  lines  without  producing  them 
to  their  point  of  intersection. 

Construction :  At  P,  any  point  in  RS,  draw 
PA  II  to  XY\  bisect  ZAPS  by  PB.  At  any 
point  in  P.B  erect  ML  _L  to  PB,  meeting  the 
given  lines  in  M  and  L.  Bisect  ML  at  D  and 
erect  DC  ±  ML,  etc. 

156.  Construct  a  common  external  tangent  to  two  circles. 
Construction:   Using  0  as  a  center  and  a 

radius  equal  to  the  difference  of  the  given 
radii,  construct  (dotted)  circle.  Draw  QA 
tangent  to  this  O  from  Q;  draw  radius  OA 
and  produce  it  to  meet  given  O  at  B.  Draw 
radius  QC II  to  OB.  Join  BC. 

Statement:  BC  is  tangent  to  both  (D. 

Proof:  AB  =  CQ  (Const.).     AB  is  II  to  CQ  (?). 

/.  ABCQ  is  a  O  (?).  But  Z  OAQ  is  a 
rt.  Z;  etc. 

157.  Construct  a  common  internal  tan- 
gent to  two  circles. 

Construction :  Using  0  as  a  center  and 
a  radius  equal  to  the  sum  of  the  given 
radii,  construct  (dotted)  circle.  Draw  QA 
tangent  to  this  0  from  Q.  Draw  radius 
OA  meeting  given  O  at  B,  etc.,  as  above. 


BOOK  III 

PROPORTION.     SIMILAR  FIGURES 

274.  A  ratio  is  the  quotient  of  one  quantity  divided  by 
another,  both  being  of  the  same  kind. 

275.  A  proportion  is  an  equation  whose  members  are  ratios. 

276.  The  extremes  of  a  proportion  are  the  first  and  the 
last  terms.     The  means  of  a  proportion  are  the  second  and 
the  third  terms. 

277.  The  antecedents  are  the  first  and  the  third  terms. 
The  consequents  are  the  second  and  the  fourth  terms. 

278.  A  mean  proportional  is  the  second  or  the  third  term  of 
a  proportion  in  which  the  means  are  identical. 

A  third  proportional  is  the  last  term  of  a  proportion  in 
which  the  means  are  identical. 

A  fourth  proportional  is  the  last  term  of  a  proportion  in 
which  the  means  are  not  identical. 

279.  A  series  of  equal  ratios  is  the  equality  of  more  than 
two  ratios. 

A  continued  proportion  is  a  series  of  equal  ratios  in  which 
the  consequent  of  any  ratio  is  the  antecedent  of  the  next 
following  ratio. 

EXPLANATORY.     A  ratio  is  written  as  a  fraction  or  as  an  indicated 

division  ;  - ,  or  a  -f-  b,  or  a :  b.     A  proportion  is  usually  written  -  —  ?,  or 
b  by 

a:b  =  x:y,  and  is  read :  "  a  is  to  b  as  x  is  to  y"  In  this  proportion  the 
extremes  are  a  and  y\  the  means  are  b  and  x ;  the  antecedents  are  a  and 
a:;  the  consequents  are  b  and  #;  and  y  is  a  fourth  proportional  to  a,  b,  x. 
In  the  proportion  a  :  m  =  m :  z,  the  mean  proportional  is  m,  and  z  is  the 
third  proportional. 

143 


144  BOOK  III.     PLANE   GEOMETRY 

THEOREMS  AND  DEMONSTRATIONS 
PROPOSITION  I.    THEOREM 

280.  In  a  proportion  the  product  of  the  extremes  is  equal  to 
the  product  of  the  means. 

Given :  |  =  -  or  a :  b  =  x :  y.     To  Prove :  ay  =  bx. 
& 

Proof:   7  =  -  (Hyp.).     Multiply  by  the  common  denomi- 

t/ 
nator  by  and  obtain,  ay  =  bx  (Ax.  3). 

Q.E.D. 

PROPOSITION  II.     THEOREM 

281.  If  the  product  of  two  quantities  is  equal  to  the  product 
of  two  others,  one  pair  may  be  made  the  extremes  of  a  propor- 
tion and  the  other  pair  the  means. 

Given :  ay  =  bx. 

To  Prove :  These  eight  proportions : 

1.  a  :  b  =  x  :  y,  5.  x  :  y  =  a  :  b, 

2.  a :  x  =  b  :  y,  6.  x  :  a  =  y  i  b, 

3.  b  :  a  =  y  :  a/,  7.  y  :  x  =  b  :  a, 

4.  b  :  y  =  a  :  x,  8.  y  :  b  =  x  :  a. 

Proof.  1.  ay  =  lx  (Hyp.). 

Divide  each  member  by  5v,    —  =  —  •  (Ax.  3). 

by      by 

.•.  j-  =  -  QY  a:b  =  xiy  Q.E.D. 

2.  ay  =  bx  (Hyp.). 
Divide  by  xy,  etc. 

3.  ay  =  bx  (?). 
Divide  by  ax,  etc.  Etc.,  etc. 

NUMERICAL  ILLUSTRATION.  Suppose  in  this  paragraph  a  =  4,  b  =  14, 
x  =  6,  y  =  21 ;  the  truth  of  the  above  proportions  can  be  clearly  seen  by 
writing  these  equivalents.  4  x  21  =  14  x  6  (True). 

1.   4  : 14  =  6  :  21  (True)  ;  2.    4  :  6  =  14  :  21  (True)  ;  etc. 

They  will  all  be  recognized  as  true  proportions. 


THEORY  OF  PROPORTION  145 

PROPOSITION  III.     THEOREM 

282.  In  any  proportion  the  terms  are  also  in  proportion  by 
alternation  (that  is,  the  first  term  is  to  the  third  as  the  sec- 
ond is  to  the  fourth). 

Given  :  a  :  b  =  x  :  y.    To  Prove  :  a:x=b:y. 
Proof  :  a:b  =  xiy  (Hyp.). 

.:ay  =  bx  (280). 

.:a:x=b:i/  (281). 

Q.E.D. 

PROPOSITION  IV.     THEOREM 

283.  In  any  proportion  the  terms  are  also  in  proportion  by 
inversion  (that  is,  the  second  term  is  to  the  first  as  the  fourth 
term  is  to  the  third). 

[The  proof  is  similar  to  the  proof  of  282.] 

PROPOSITION  V.     THEOREM 

284.  In  any  proportion  the  terms  are  also  in  proportion  by 
composition  (that  is,  the  sum  of  the  first  two  terms  is  to  the 
first,  or  the  second,  as  the  sum  of  the  last  two  terms  is  to 
the  third,  or  the  fourth). 

•  I  a  +  b  :  a=x+  V  :  X,  or 

Given:  a:b  =  x:y.     To  Prove: 


Proof  :  a:b  =  x:y  (Hyp.).     .  -.  ay  =  bx  (280). 

Add  ax  to  each,  and  obtain,  ax  +  ay  =  ax  +  bx  (Ax.  2). 
That  is,                   a  (x  +  y)  =  x(a  +  b). 

Hence                       a+b  :  a  =  x  +  y  :  x  (281). 

Similarly,  by  adding  by,  a  +  b  :  b  =  x  -f  y  :  y.  Q.E.D. 


Ex.  1.  Is-the  equation  12  :  9  =  28  : 21  a  true  proportion  ? 

Ex.  2.  Apply  Proposition  I  to  the  above  proportion. 

Ex.  3.  Apply  Proposition  III  to  the  above  proportion. 

Ex.  4.  Apply  Proposition  IV  to  the  above  proportion. 

Ex.  6.  Apply  Proposition  V  to  the  above  proportion. 


146  BOOK  III.     PLANE   GEOMETRY 

PROPOSITION  VI.    THEOREM 

285.  In  any  proportion  the  terms  are  also  in  proportion  by 
division  (that  is,  the  difference  between  the  first  two  terms 
is  to  the  first,  or  the  second,  as  the  difference  between  the 
last"  two  terms  is  to  the  third,  or  the  foiirth). 

Given:  a:  b  =  x  :y.     ToProve:   {«-»'«-*-»  =  «.« 

a  —  o  :  o  =  x  —  y  -  y> 

Proof :  a:b  =  x:y  (Hyp.).     .  •.  ay  =  bx  (280). 

Subtracting  each  side  from  ax,  ax—  ay  =  ax  —  bx  (Ax.  2). 
That  is,                 a(x  —  y)  =  x(a  —  b). 

Hence                     a—'b-.a  —  x  —  y-.x  (281). 

Similarly,               a  —  b:b  =  x  —  y:y.  Q.E.D. 


NOTE  1.     The  proportions  of  284  and  285  may  be  written  in  many 
different  forms  (282,  283)  .     Thus,  (1)  a  ±b:a  =  x  ±y\x\ 
(2)  a±b:b  =  x±y:y;  (3)  a±b:x±y  =  a:x,  etc. 

NOTE  2.  In  any  proportion  the  sum  of  the  antecedents  is  to  the 
sum  of  the  consequents  as  either  antecedent  is  to  its  consequent.  Also, 
in  any  proportion  the  difference  of  the  antecedents  is  to  the  difference  of 
the  consequents  as  either  antecedent  is  to  its  consequent.  (Explain.) 

Thus  :  a  +  x:b  +  y  =  a:b  =  x:y.     Also,  a  —  x:b  —y  =  a:b  =  x:y. 

PROPOSITION  VII.     THEOREM 

286.  In  any  proportion  the  terms  are  also  in  proportion  by 
composition  and  division  (that  is,  the  sum  of  the  first  two 
terms  is  to  their  difference  as  the  sum  of  the  last  two  terms 
is  to  their  difference). 

Given:  a:b=x:y.     ToProve:  ^±-^  =  ^±1. 

a  —  b     x—  y 

Proof:  fL±i  =  ^±^  (284). 


Divide  the  first  by  the  second,  ^±-   =  ?-l  (Ax.  3). 

••7*  x-y       Q.E.D. 


THEORY   OF  PROPORTION  147 

PROPOSITION  VIII.     THEOREM 

287.  In  any  proportion,  like  powers  of  the  terms  are  also  in 
proportion,  and  like  roots  of  the  terms  are  in  proportion. 

Given  :  a  :  b  =  x  :  y. 

To  Prove  :  an  :  bn  =  xn  :  yn  ;  and  -\/a  :  -\/b  =  ^/x  :  ^Jy. 

Proof:  [Write  the  hypothesis  in  fractional  form,  etc.] 

PROPOSITION  IX.     THEOREM 

288.  In  two  or  more  proportions  the  products  of  the  corre- 
sponding terms  are  also  in  proportion. 

Given  :  a\~b  —  x\y^  and  c  :  d  =  I  :  m,  and  e  :  f  =  r  :  s. 

To  Prove  :  ace  :  bdf  =  xlr  :  yms. 

Proof:  [Write  in  fractional  form  and  multiply.] 

PROPOSITION  X.     THEOREM 

289.  A  mean  proportional  is  equal  to  the  square  root  of  the 
product  of  the  extremes. 

Given  :  a  :  x  =  x  :  b.     To  Prove  :  x  =  Va6. 
Proof:  [Use  280.] 

PROPOSITION  XI.     THEOREM 

290.  If  three  terms  of  one  proportion  are  equal  to  the  corre- 
sponding three  terms  of  another  proportion,  each  to  each,  the 
remaining  terms  are  also  equal. 


Given:  '  .     ToProve:m  =  r. 

\  a  :  b  =  c  :  r. 

Proof  :  am  =  be  and  ar  =  be  (280). 

.•.am  =  ar  (Ax.  1). 

.-.  m  =  r  (Ax.  3). 

Q.E.D. 


148  BOOK  III.     PLANE   GEOMETRY 

PROPOSITION  XII.     THEOREM 

291.  In  a  series  of  equal  ratios  the  sum  of  all  the  ante- 
cedents is  to  the  sum  of  all  the  consequents  as  any  antecedent 
is  to  its  consequent. 

Given:  f=£  =  f  =  f- 
b      d     f     h 

To  Prove:  *  +  '  +  e  +  f  =  f=^  =  etc. 
b+d+f+h      b      d 

Proof  :  Set  each  given  ratio  =  to  m  ;  thus, 

a  c  e  a 

-  =  m;   3  =  w  ;   -r  =  w  ;   ^  =  m. 

b  d  f  h 

.  •.  a  =  bm,  c  =  dm,  e  =/m,  g  =  Jim  (Ax.  3). 

TT  a  +  £+£+</      bm+  dm+fm  +  hm,  ,0   t    ...    ,.     N 

Hence,       '       '         *  =  -  f  --  ^     ^"  -  (Substitution). 


(Canceling). 


Ex.l.   If  3:4  =  6:a:,finda:.  Ex.2.   If  8  : 12  =  12  :  *,  find  x. 

Ex.  3.   Find  a  fourth  proportional  to  6,  7,  and  15. 

Ex.  4.    Find  a  third  proportional  to  4  and  10. 

Ex.  6.   If  11  : 15  =  x  :  25,  find  x.         Ex.  6.   If  4  :  x  =  x  :  25,  find  x. 

Ex.  7.   Find  a  mean  proportional  between  8  and  18. 

Ex.  8.   If  7  :  x  =  35  :  48,  find  x. 

Ex.  9.  Given,  5:8=  15  : 24.  Write  seven  other  true  proportions  con- 
taining these  four  numbers. 

Ex.  10.   If  5  X  6  =  2  x  15,  write  eight  proportions  with  these  numbers. 

Ex.  11.  If  7  : 12  =  21 :  36,  write  the  proportion  resulting  by  alternation ; 
inversion;  composition;  division;  composition  and  division. 

Ex.  12.  If  x  +  y  :  x  —  y  =  17  : 7,  write  the  proportions  that  result  by 
virtue  of  composition ;  division  ;  composition  and  division. 

Ex.  13.    Apply  291  to  the  ratios,  f  =  f  =  |  =  1^  =  ^n. 


THEORY   OE  PROPORTION  149 

292.  A  segment  of  a  line  is  any  part  of  the  line.  If  a  line 
is  divided  by  a  point  into  two  segments,  they  are  the  distances 
from  this  point  of  division  to  the  extremities  of  the  line. 

The  upper  line  AB  is  divided  internally  by  P 

(a  point  between  the  extremities  A  and  B}  into  .  P     B 

segments  AP  and  PB.     The  lower  line  AB  is 

divided  externally  by  R  (a  point  in  the  prolonga-      R A  Q 

tion  of  AB}  into  segments,  RA  and  RB. 

When  a  line  is  divided  internally,  it  equals  the  sum  of  the  segments; 
when  it  is  divided  externally,  it  equals  the  difference  of  the  segments. 

Two  lines  are  divided  proportionally  if  the  ratio  of  the  lines  is  equal 
to  the  ratio  of  the  corresponding  segments.    Thus,  A  C 
and  AE  are  said  to  be  "  divided  proportionally  "  if 


AC  =AB      AC  ^BC       AC=AB=BC 

AE     AD      AE     DE      AE     AD     DE  ,  - 

NOTE.     We  have  seen  that  it  is  possible  to  add 

two  lines  and  subtract  one  line  from  another.  Now  it  is  essential  that 
we  clearly  understand  the  significance  implied  by  indicating  the  multi- 
plication or  the  division  of  one  line  by  another. 

What  is  actually  done  is  to  multiply  or  divide  the  numerical  measure 
of  one  line  by  the  numerical  measure  of  another.  Thus,  if  one  line  is  8 
inches  long  and  another  is  18  inches  long,  we  say  that  the  ratio  of  the 
first  line  to  the  second  is  fa  or  f,  meaning  that  the  smaller  line  is  four 
ninths  of  the  larger. 

Also,  in  referring  to  the  product  of  two  lines,  we  merely  understand 
that  the  product  of  their  numerical  measures  is  intended. 

If  a  line  is  multiplied  by  itself,  we  obtain  the  square  of  the  numerical 
measure  of  the  line.  The  square  of  the  line  AB  is  written  AB2  or 
(AB)2,  and  the  quantity  that  is  squared  is  the  numerical  value  of  the 
length  of  AB. 

In  the  preceding  paragraphs  of  Book  III,  we  have  been  considering 
numerical  magnitudes.  It  should  be  distinctly  understood  that  in  the 
following  geometrical  propositions  and  demonstrations,  the  foregoing 
interpretation  is  implied  in  multiplication  and  division  involving  lines. 

Historical  Note.  Eudoxus,  one  of  the  most  prominent  of  the  Greek 
mathematicians,  was  famous  also  as  a  physician  in  the  fourth  century 
B.C.  One  of  his  principal  contributions  to  geometry  was  the  perfection 
of  a  rigorous  theory  of  ratio  and  proportion. 


150 


BOOK  III.  PLANE  GEOMETRY 


B 


PROPOSITION  XIII.     THEOREM 

293.  A  line  parallel  to  one  side  of  a 
triangle  divides  the  other  sides  into  pro- 
portional segments. 

Given :   A  ABC  and  line  LM  II  to  BC. 

To  Prove :  AL  :  LB  =  AM  :  MC. 

Proof :   I.   If  the  parts  AL  and  LB  are  commensurable. 

There  exists  a  common  unit  of  measure  of  AL  and  LB  (225). 
Suppose  this  is  contained  9  times  in  AL  and  5  times  in  LB. 

Then  ^=|  (Ax.  3). 

LB      5 

Draw  lines  through  the  several  points  of  division  II  to  BC. 
These  divide  AM  into  9  parts  and  MC  into  5  parts. 

All  these  14  parts  are  equal  (140). 


MC 
AL 


AM 


II.    If  the  parts  AL  and  LB  are  incom- 
mensurable. 

There  does  not  exist  a  common  unit 

(225). 
Divide  AL  into  several  equal  parts 

(by  261). 


(Ax.  3). 
(Ax.  1). 

Q.E.D. 


B 


There  is 
(225). 


Apply  one  of  these  as  a  unit  of  measure  to  LB. 
remainder,  RB 

Draw  RS  II  to  BC. 

Now  —  =  —  (Case  I). 

LR       MS 

Indefinitely  increase  the  number  of  equal  parts  of  AL.  That 
is,  indefinitely  decrease  each  part,  the  unit  or  divisor.  Hence, 
the  remainder,  EJB,  is  indefinitely  decreased.  (Because  the 
remainder  is  <  the  divisor.) 


PROPORTIONAL   LINES  151 

That  is,  RB  approaches  zero  as  a  limit. 
Also  8C  approaches  zero  as  a  limit. 

.  •.  LR  approaches  LB  as  a  limit  (227). 

Also  MS  approaches  MC  as  a  limit  (227). 

AL  AL  ,.     ., 

.-.  —  approaches  —  as  a  limit. 
LR  LB 

AM  AM  v     ., 

Also  -  approaches  -   -  as  a  limit. 

MS  MC 

Consequently,  —  =  —  (229).     Q.B.D. 

LB       MC 

PROPOSITION  XIV.     THEOREM 

294.   If  a  line  parallel  to  one  side  of  a  triangle  intersects  the 
other  sides,  it  divides  these  sides  propor-         A 
tionally. 

Given:   A  ABC-,  LM  II  to  BC. 

To  Prove  : 


I.  AB  i  AC  =  AL  :  AM.  / 


II.  AB:AC  =  LB:MC.  B 

Proof  :  AL  :  LB  =  AM  :  MC  (293)  . 

.'.  AL  -{-LB:  AL  =  AM  +  MC  :  AM  (284). 

Also  AL  +  LB  :  LB  =  AM  +  MC  :  MC  (284). 

But  AL  +  LB  =  AB,  and  AM  +  MC  =  AC  (Ax.  4). 

Substituting,  in  last  two  proportions  : 

AB  :  AL  =  AC  :  AM  (Ax.  6). 

Also  AB:  LB  =  AC:  MC  (Ax.  6). 

.  •.  I.  AB  :  AC  =  AL  :  AM  (282). 

Also  II.  AB  :  AC=  LB  :  MC  (282).      Q.E.D. 

These  proportions  may  be  combined  thus  :  —  =  —  =  -- 

AC      AM     MC 

Each  of  the  above  proportions  may  be  written  in  eight  dif- 
ferent ways. 

Ex.    If,  in  figure  of  294,  AB  is  9  units,  AC  12  units,  and  AL  6  units, 
find  AM,  LB,  and  MC. 


152  BOOK  III.     PLANE   GEOMETKY 

PROPOSITION  XV.     THEOREM 

295.  Three  or  more  parallels  intercept 
proportional  segments  on  two  transversals. 

Given:  (?). 
To  Prove : 

AC  :  BD  =  CE  :  DF  =  EG  :  FH.  f  ^ 

Proof:  Draw  from  A,  AT\\  to  BH  intersecting  the  Us,  etc. 
InAAES,  AE==AC^=CE  294 

A8       AE       ES 

InAAGT,  —  = —  (294). 

AS       ST 

AC       CE      EG  ^A        IN 

.'. = =  -  (Ax.  1). 

AE       ES      ST 

But  AE  =  BD,   BS  =  DF,    ST  =  FH  (124). 

Hence  A  C :  BD  =  CE :  DF  =  EG  :  FH  (Ax.  6) . 

Q.E.D. 

PROPOSITION  XVI.     THEOREM 

296.  If  a  line  divides  two  sides  of   a 
triangle  proportionally,  it  is  parallel  to  the 
third  side. 

Given:   A  ABC;   line  DE;    the  propor- 
tion AB  :  AC  =  AD :  AE. 
To  Prove :  DE  is  II  to  BC. 
Proof:  Through  D  draw  DX  II  to  BC  meeting  AC  at  X. 

.:AB:AC=AD:AX  (294). 

But                                 AB:AC=  AD'.AE  (Hyp.). 

.-.  AX=AE  (290). 

.•.  DX  and  DE  coincide  (39). 

That  is,                        DE  is  II  to  BC  Q.E.D. 

Ex.  1.  Prove,  as  296  is  proved,  that  the  line  bisecting  two  sides  of  a 
triangle  is  parallel  to  the  third  side. 

Ex.  2.  In  fig.  of  Proposition  XIV,  if  AL  is  $f  of  AB,  what  is  true  of 
AMI  of 


PROPORTIONAL  LINES  153 

PROPOSITION  XVII.     THEOREM 

297.  The  bisector  of  an  angle  of  a  triangle  divides  the  oppo- 
site side  into  segments  that  are  propor-     p,. 

tional  to  the  other  two  sides.  '' 

Given:    A  ABC;    BS  the  bisector   of 
Z  ABC. 

To  Prove:  AS:SC=AB:BC.  A        5 

Proof:    Through   A   draw   AP  II  to  BS,  meeting  CB,  pro- 
duced, at  P. 

Then,  in  A  PAC,        AS :  SC  =  PB:BG  (294) . 

Now  Zm  =  Zz  (67). 

And  Zw  =  Zz  (66). 

But  Z  m  =  Z  n  (Hyp.). 

.-.  Ax=  Z.z  (Ax.  1). 

.-.  PB  =  AB  (H4). 

Substituting  above,  AS:  SC  =  AB  :  BC       (Ax.  6).     Q.E.D. 

PROPOSITION  XVIII.     THEOREM 

298.  The  bisector  of  an  exterior  angle  of  a  triangle  divides 
the  opposite  side  (externally)  into  segments  that  are  propor- 
tional to  the  other  two  sides. 

Given:  A  ABC;  BSf,  the  bisector  of 
exterior    /.  ABD,    meeting   AC   (exter- 

nally)  at  s'.  

To  Prove :  8fA  :  s'C  =  AB :  BC. 

Proof :  Through  A  draw  AP  II  to  BS1  meeting  BC  at  P. 

Then,  mACBS',        SfA:SfC=BP:BC  (294). 

Now                                   Zra=Zz  (67). 

And                                    Zn  =  Z.z  (66). 

But                                   Z  m  =  Z  n  (Hyp.). 

.:^x  =  ^z  (Ax.  1). 

.-.  BP  =  AB  (114). 

Substituting  above,  SfA  :  sfc  =  AB  :  BC  (Ax.  6).     Q.E.D. 

ROBBINS'S    NEW    PLANE    GEOM.  11 


154 


BOOK   III.     PLANE   GEOMETRY 


Ex.  1.  If,  in  297,  AS  =  3,  AB  =  4,  EC  =  9,  find  SC. 

Ex.  2.  If,  in  297,  AC  =  20,  AB  =  9,  BC  =  21,  find  .45  and  SC. 

Ex.  3.  If,  in  298,  S'A  =  10,  AB  =  7,BC  =  16,  find  S'C  and  ylC. 

Ex.  4.  If,  in  298,  .4  C  =  14,  4J5  =  12,  £C  =  19,  find  S'.-l  and  S'C. 

299.  A  line  is  divided  harmonically  if   it   is   divided  in- 
ternally and  externally  in  the  same  ratio. 

In  297,  the  line  A  C  is  divided  internally  by  5,  in  the  ratio  AB:BC. 
In  298,  the  line  ^4C  is  divided  externally  by  S',  in  the  ratio  AB:BC. 

PROPOSITION  XIX.     THEOREM 

300.  The  bisectors  of  the  interior  and  exterior  angles  of  a 
triangle  (at  a  vertex)  divide  the  opposite 

side  harmonically.  - — B 

Given:  A  ABC;  BS  bisecting  Z  ABC; 
and  BSf  bisecting  Z  ABD. 

To  Prove :  AS  :  sc  =  S'A  :  s'c. 

AS^AB 
SC~  BC 

S'A  _  AB 

~S*C~  BC 
AS  _8'A 

'  '  sc~  s'c 


S' 


Proof: 


(297). 
(298). 


(Ax.  1). 

Q.E.D. 

301.  Similar  polygons  are  polygons  that  are  mutually 
equiangular  and  the  homologous  sides  of  which  are  pro- 
portional. That  is,  every  pair  of  homolo- 
gous angles  are  equal ;  and  the  ratio  of 
one  pair  of  homologous  sides  is  equal  to 
the  ratio  of  every  other  pair  of  homolo- 
gous sides, 

a  :  a'  =  b  :  V  =  c  :  cr  =  d  :  d'  =  etc. 

Triangles  are  similar  if  they  are  mutually  equiangular  and 
their  homologous  sides  are  proportional. 


SIMILAR  POLYGONS  155 

PROPOSITION  XX.     THEOREM 

302.    Two  triangles  are  similar  if  they  are  mutually  equi- 
angular. 

Given :  A  ABC,  DEF\  Z^  =  ZD,  Z  £  =  Z  #,  Z  c  =  Z  2/1 
To  Prove :  The  A  are  similar  (that  is,  their  sides  are  pro- 
portional). 


Proof :  Place  A  ABC  upon  A  DEF  so  that  Z  A  coincides 
with  its  equal,  Z  D,  and  A  A  EC  takes  the  position  of  A  DBS. 
Then  Z  DBS  =  Z  E  (Hyp.). 

.-.  BS  is  II  to  EF  (71). 

/.  DE:DB  =  DF:D8  (294). 

That  is,  DE:AB  =  DF:  AC  (Ax.  6). 

Likewise,  by  placing  A  ABC  upon  A  DEF  so  that  Z  B  coin- 
cides with  its  equal,  Z  E,  we  may  prove  that 

DE:AB  =  EF:BC 

.'.  DE:AB=  DF:AC=EF:BC  (Ax.  1). 

.-.  the  A  are  similar  (301). 

Q  E.D. 

303.  COROLLARY.    Two  triangles  are  similar  if  two  angles 
of  one  are  equal  to  two  angles  of  the  other.  (Ill  and  302.) 

304.  COROLLARY.    Two  right  triangles  are  similar  if  an  acute 
angle  of  one  is  equal  to  an  acute  angle  of  the  other.  (303.) 


Ex.  1.  If  two  transversals  intersect  between  two  parallels,  the  tri- 
angles formed  are  similar. 

Ex.  2.  Two  isosceles  triangles  are  similar  if  a  base  angle  of  one  is 
equal  to  a  base  angle  of  the  other. 


156 


BOOK   III.     PLANE   GEOMETRY 


through  P  three  lines 
A 


Ex.  3.    Two  isosceles  triangles  are  similar  if  the  vertex  angle  of  one 
is  equal  to  the  vertex  angle  of  the  other. 

Ex.  4.   The  line  joining  the  midpoints  of  two  sides  of  a  triangle  forms 
a  triangle  similar  to  the  original  triangle. 

Ex.  5.    The  diagonals  of  a  trapezoid  form,  with  the  parallel  sides,  two 
similar  triangles. 

Ex.  6.   If  at  the  extremities   of  the  hypotenuse  of  a  right   triangle 
perpendiculars  are  erected  meeting  the  legs  produced,  the 
new  triangles  formed  are  similar. 

Ex.  7.   In  the  figure  of  Ex.  6,  prove : 

(1)  Triangle  ABC  similar  to  each  of  the  triangles 
ACE  and  BCD. 

(2)  Triangle  ABE  similar  to  triangle  ABD. 

(3)  Triangle  A  CE  similar  to  triangle  ABD. 

(4)  Triangle  BCD  similar  to  triangle  ABE. 

(5)  Triangles  ABC,  ABD,  ABE  similar. 

Ex.  8.    Two  circles  are  tangent  externally  at  P : 
are  drawn,  meeting  one  circumference  in  A, 
B,  C,  and  the  other  in  A',  B',  C'.     The  tri- 
angles ABC  and  A'B'C'  are  similar. 

Ex.  9.   Prove  the  same  theorem  if  the  cir- 
cles are  tangent  internally. 

Ex.  10.   If  two  circles  are  tangent  externally 
at  P,  and  BB1,  CC'  are  drawn  through  P,  terminating  in  the  circum- 
ferences, the  triangles  PEC  and  PB'C'  are  similar. 

[Draw  the  common  tangent  at  P.] 

Ex.  11.   Prove  the  same  theorem  if  the  circles  are  tangent  internally. 

Ex.  12.   If  AD  and  BE  are   two  altitudes  of  A 

triangle  ABC,  the  triangles  A  CD  and  BCE  are 
similar. 

Ex.  13.  If  AD  and  BE  are  two  altitudes  of 
triangle  ABC,  meeting  at  0,  the  triangles  BOD 
and  A  OE  are  similar. 

Ex.  14.  Triangle  ABC  is  inscribed  in  a  circle  and 
AP  is  drawn  to  P,  the  midpoint  of  arc  BC,  meeting 
chord  CB  at  D.  The  triangles  ABD  and  ACP  are 
similar. 


SIMILAR   POLYGONS  157 

PROPOSITION  XXI.     THEOREM 

305.  If  a  line  parallel  to  one  side  of  A 
a  triangle  intersects  the  other  sides,  the 
triangle  formed  is  similar  to  the  original 
triangle. 

Given :  MN  II  to  BC  in  A  ABC. 

To  Prove :  A  AMN  similar  to  A  ABC.       B~~  ~~C 

Proof:  In  the  A  AMN  and  ABC, 

/.A  =  /-A  (Ident.). 

ZAMN=ZB  (67). 

.-.  A  AMN  is  similar  to  A  ABO  (303). 

Q.E.D. 

PROPOSITION  XXII.     THEOREM 

306.  If  two  triangles  have  an  angle  of  one  equal  to  an  angle 
of  the  other  and  the  sides  including  these  angles  proportional, 
the  triangles  are  similar. 


s 


/  \ 


B  C      E  F 

Given:  A  ABC  and  DEF-,  Z.A  =  Z.D-,  DE :  AB  =  I>F :  AC. 
To  Prove :  The  A  similar. 

Proof :  Superpose  A  ABC  upon  A  DEF  so  that  Z.  A  coincides 

with  its  equal,  Z  D,  and  A  ABC  takes  the  position  of  A  DBS. 

Now  DE:DB  =  DF:DS  (Hyp.). 

.-.  US  is  II  to  EF  (296). 

.-.  A  DBS  is  similar  to  A  DEF  (305). 

Substituting,  A  ABC  is  similar  to  A  DEF  (Ax.  6). 

Q.E.D. 


158  BOOK   III.    PLANE   GEOMETRY 

307.  COROLLARY.     If  two  triangles  are  similar  to  the  same 
triangle,  they  are  similar  to  each  other. 

Proof:  The  three  angles  of  each  of  the  first  two  triangles 

are  respectively  equal  to  the  three  angles  of  the  third  (301). 

Hence  the  first  two  A  are  mutually  equiangular    (Ax.  1). 

Therefore  they  are  similar  (302). 

Q.E.D. 

PROPOSITION  XXIII.     THEOREM 

308.  If  two  triangles  have  their  homologous  sides  propor- 
tional) they  are  similar. 

D 


B  C      E  F 

Given :  A  ABC  and  DEF\  DE  :  AB  =  DF  :  AC  =  EF  :  EC. 
To  Prove :  A  ABC  similar  to  A  DEF. 

Proof :  On  DE  take  DK  =  to  AB  ;  and  on  DF  take  DL  =  to 
AC.     Draw  KL. 

1st    Now  DE  :  AB  =  DF  :  AC  (Hyp.). 

DE  :  DK  =  DF  :  DL  (Ax.  6). 

.-.  EL  is  11  to  EF  (296). 

A  DKL  is  similar  to  A  DEF  (305). 

2d  DE  :  DK  =  EF  :  KL  (Def.  sim.  A)  (301). 

Substituting,  DE :  AB  =  EF :  KL  (Ax.  6). 

But  DE:AB  =  EF:BC  (Hyp.). 

.'.BC  =  KL  (290). 

3d  A  ABC  is  congruent  to  A  DKL  (78). 

But  A  DKL  has  been  proved  similar  to  A  DEF. 
Substituting,     A  ABC  is  similar  to  A  DEF  (Ax.  6). 

Q.E.D. 


SIMILAR  POLYGONS  159 

PROPOSITION  XXIV.     THEOREM 

309.  If  two  triangles  have  their  homologous  sides  parallel, 
they  are  similar. 

A  D 


E  F 

Given :  A  ABO  and  DEF ;  AB  II  to  DE  ;  AC  II  to  DF ;  and  BC 

II  to  EF. 

To  Prove :  A  ABC  similar  to  A  DEF. 

Proof :  Produce  BC  of  A  ABC  until  it  intersects  two  sides 
of  A  DEF  at  E  and  s. 

Now  Z  B  =  Z  DBS,  and  Z  DBS  =  Z  E  (67). 

.-.  ZB  =  /.E  (Ax.  1). 

Similarly,  we  may  prove  Z  ACB  =  to  Z  F. 

.  •.  A  ^BC  is  similar  to  A  DEF  (303). 

Q.E.D. 

Ex.1.    Draw  the  figure  and  prove  Proposition  XXIV: 

(a)  If  one  triangle  is  entirely  within  the  other ; 

(6)  If  no  side  of  either,  when  prolonged,  meets  any  side  of  the  other ; 

(c)  If  one  side  of  one  intersects  two  sides  of  the  other,  without  pro- 
longation ; 

(d)  If  a  vertex  of  one  is  upon  a  side  of  the  other. 

Ex.  2.  If  in  a  right  triangle,  a  perpendicular  is  drawn  from  the  vertex 
of  the  right  angle  upon  the  hypotenuse,  the  two  new  right  triangles  are 
similar  to  the  original  triangle  and  to  each  other. 

Ex.  3.    Are  all  equilateral  triangles  similar?    Why? 

Ex.  4.    Are  all  squares  similar?  all  rectangles?     Why? 

Ex.  5.  Do  a  square  and  a  rectangle  fulfill  either  of  the  conditions  for 
similar  polygons? 

Ex.  6.  If  from  any  point  in  a  leg  of  a  right  triangle  a  line  is  drawn 
perpendicular  to  the  hypotenuse,  the  triangles  are  similar. 


160  BOOK   III.     PLANE   GEOMETRY 


PROPOSITION  XXV.     THEOREM 

310.  If  two  triangles  have  their  homologous  sides  perpen- 
dicular, they  are  similar. 


M 


Given:  A  ABC  and  DEF -,  AB  _L  to  VE\  AC  ±.  to  DF\  etc. 
To  Prove :  A  ABC  similar  to  A  DEF. 

Proof:  Through  P,  any  point  in  EF,  construct  PR  I!  to  AC, 
meeting  DF  at  M.     At  R,  any  point  in  PM,  draw  RS  II  to  AB, 
meeting  ED  at  N.     Draw  PS  II  to  BC,  meeting  NR  at  8,  form- 
ing the  A  PRS.     Now  PM  is  J_  to  DF  and  RN  is  _L  to  DE    (64) . 
In  quadrilateral  DMRN, 

Z  D  +  Z  3f  +  Z  MRN  +  Z  zV  =  4  rt.  /4  (156). 

But  Zjtf  + Z  TV  =  2  rt.  ^  (16). 

.-.  ZD  +/.MRN  =  2rt.  A       (Ax.  2). 

But  Z  a  +  Z  ^f##  =  2  rt.  A  (46). 

.-.  ZD-Za  (49). 

Similarly,  by  quadrilateral  ^P5JV,  it  may  be  proved  that 

Z#  =  Z6. 

.  •.  A  DEF  is  similar  to  A  pis  (303). 

But  A  ABC  is  similar  to  A  PRS  (309). 

.  •.  A  ABC  is  similar  to  A  DEF  (307). 

Q.E.D. 

Historical  Note.  Thales,  as  early  as  the  sixth  century  B.C.,  used  simi- 
lar triangles  to  determine  the  height  of  the  great  Egyptian  pyramid.  He 
measured  the  shadow  of  a  pole  of  known  height,  and  the  shadow  of  the 
pyramid  at  the  same  time,  to  obtain  homologous  sides  of  similar  right 
triangles. 


SIMILAR   POLYGONS 


161 


PROPOSITION  XXVI.     THEOREM 

311.   Two  homologous  altitudes  of  two  similar  triangles  are 
proportional  to  any  two  homologous  sides. 
Given:  (?). 
To  Prove  • 


BL 


AB 


AC 


BC 


B'L'     A'B'     A'C'     B'C' 


A  L      C     A' 

Proof :   A  ABC  is  similar  to  A  A' B'C' 

.:  A  ABL  is  similar  to  A  A1  B'L' 
BL         AB 


C 


But 


BL 


B'L' 
AB 

A'B' 
AB 


BC 


A'B' 

AC  = 

A'C'~  B'C' 

AC         BC 


A'B'     A'C'     B'C' 


(Hyp.)- 
(301). 
(304). 

(301). 

(301). 

(Ax.  1). 

Q.E.D. 


312.  In  similar  figures,  homologous  angles  are  equal  (Def.). 

313.  In  similar  figures,  homologous  sides  are  proportional 

(Def.). 

The  antecedents  of  this  proportion  belong  to  one  of  the  similar  figures, 
and  the  consequents  to  the  other. 

314.  In  similar  triangles,  homologous  sides  are  opposite 
homologous  angles. 

Shortest  sides  are  homologous.  (Opposite  smallest  A}. 
Medium  sides  are  homologous.  (Opposite  medium  A}. 
Longest  sides  are  homologous.  (Opposite  largest  A). 


162 


BOOK   III.     PLANE   GEOMETRY 


PROPOSITION  XXVII.     THEOREM 
315.   If  two  parallel  lines  are  cut  by 
three  or  more  transversals  that  meet  at 
a  point,  the  corresponding  segments  of  the 
parallels  are  proportional. 
Given:  (?). 
To  Prove:   ^=^  =  **. 

CG       GH       HD 
Proof :   In  A  OCG,  AE  is  II  to  CG 
.'.  A  OAE  is  similar  to  A  OCG 
Likewise,  A  OEF  is  similar  to  A  OGH,  and  A  OFB 
to  A  OHD 


L     \\ 


G 


CG       OG  GH       OG 

AE  _EF 

'  CG  ~  GH 

Likewise  =  — ,  and  —  =  — 

GH       OH  OH       HD 

AE  =  EF  ^FB 
'  CG       GH~  HD 

PROPOSITION  XXVIII.     THEOREM 

316.  If  three  or  more  non-parallel  trans- 
versals intercept  proportional  segments  on 
two  parallels,  they  meet  at  a  point.  [Con- 
verse.] 

Given:  Transversals  AB,  CD,  EF;  par- 
allels AE  and  BF;  proportion,  AC:  BD  = 
CE :  DF. 

To  Prove :  AB,  CD,  EF  meet  at  a  point. 


H  D 

(Hyp.)- 

(305). 

is  similar 

(305). 

(313). 
(Ax.  1). 

(313). 
(Ax.  1). 

Q.E.D. 


kC 


B 


Proof :   Produce  AB  and  CD  until  they  meet,  at  o. 

Draw  OF  cutting  AE  at  X. 

Now  AC:BD=CX:DF  (315). 

But  AC:BD=CE:DF  (Hyp.)- 


SIMILAR  POLYGONS  163 

.-.  CX=CE  (290). 

.-.  FE  and  FX  coincide  (39). 

That  is,        FE  prolonged,  passes  through  o.  Q.E.D. 


Ex.  1.    Show  the    truth  of    Proposition    XXVIII,   by  an    accurate 
diagram. 

Ex.  2.   In  the  figure  of  Proposition  XXVIII,  what  is  point  XI 

PROPOSITION  XXIX.     THEOREM 

317.    The  perimeters  of  two  similar  polygons  are  to  each 
other  as  any  two  homologous  sides. 

D' 
D 


Given:   Polygon    R    whose    perimeter  =  P    and    similar 
polygon  8  whose  perimeter  =  Pf. 
To  Prove :  p  :  p'  =  AB  :  A'B'  =  BC  :  B'C'  =  etc. 
Proof :  AB  :  A'B'  =  BC  :  B'C'  =  CD  :  C'D'  =  etc.  (313). 

.         AB  +  BC+CD+  ...       =  AB    =   BC   =         ^          .^ 

'  A'B' +  B'C'  +  C'D'  +  ...     A'B'     B'C' 

Substituting,  y  =  —^  =  -—  =  etc.  (Ax.  6). 

Q.E.D. 


Ex.  1.  The  sides  of  a  certain  polygon  are  4,  10,  12,  15,  and  18.  The 
shortest  side  of  a  similar  polygon  is  6.  Find  the  other  four  sides. 

Ex.  2.  The  perimeters  of  two  similar  polygons  are  125  and  275 
respectively.  The  longest  side  of  the  smaller  polygon  is  40.  Find  the 
longest  side  of  the  larger  polygon. 

Ex.  3.  Enumerate  the  ways  of  proving  two  triangles  similar.  Which 
is  the  easiest  of  these  ways? 


164  BOOK   III.     PLANE   GEOMETRY 

PROPOSITION  XXX.     THEOREM 

318.  If  two  polygons  are  similar,  they  may  be  decomposed 
into  the  same  number  of  triangles,  similar  each  to  each  and 
similarly  placed. 

B' 
B 


ED  E'  D1 

Given :   Similar  polygons  BE  and  B'E'. 

To  Prove :  A  ABC  similar  to  A  A' B'C'  ; 
A  ACD  similar  to  A  A'C'D'  ; 
A  AED  similar  to  A  A'E'D' '. 

Proof:   First.     AB  :  A'B'  =  BC  :  B'C'  (313). 

Also  Z  B  =  Z  B'  (312). 

Therefore         A  ABC  is  similar  to  A  A' B'C'  (306). 

Second.     In  A  ABC  and  A' B'C',  ^-  =  -^-  (313). 

B'c'      A'C' 

T>/"f  /^7~i 

In  the  similar  polygons,  —^  =  -7-7  (313). 

B  C        CD 

Consequently  —-—=——-  (Ax.  1). 

A   C1         C1  D 

In  the  polygons,  Z  BCD  =  Z  5'c'D'  1 

In  the  A  ^IBC  and  A'B'C',  Z.  BCA=  ^  B' C* A1  } 
Hence,  by  subtraction,        Z  ACD  =  Z.  A'C'D'  (Ax.  2). 

Therefore  A  ACD  is  similar  to  A  A'C'D'  (306). 

Third.     A  AED    is    proved    similar    to    A  A'E'D'   in    like 
manner.  Q.E.D. 


Ex.    A  map  is  drawn  on  a  scale  of  1  in.  to  400  miles.     How  far  is 
Paris  from  Antwerp  if  they  are  1|  in.  apart  on  the  map? 


SIMILAR   POLYGONS 


165 


PROPOSITION  XXXI.     THEOREM 

319.  If  two  polygons  are  composed  of  triangles  similar  each 
to  each  and  similarly  placed,  the  polygons  are  similar.  [Con- 
verse.] 

H' 

H 


J  K' 

Given:   A  GHI  similar  to  A  G'H'I'  ; 

A  GIJ  similar  to  A  G'I'J'  ; 

A  GJK  similar  to  A  G'J'K'. 
To  Prove :   The  polygons  HK  and  H' K'  similar. 
Proof :    First.     In  A  HGI  and  H' G' l' ,  Z  H  =  Z  H'        (312). 
Also  in  these  A         Z  HIG  =  Z  H'I'G'  (312). 

In  A  GIJ  and  G'I'J',  Z  GIJ  =  Z  G'I'J'  (312). 

Adding,  ZniJ=^HfIfJf  (Ax.  2). 

Likewise  Z  u^r  =  Z  I'J'K'  ;  etc. 

That  is,  the  polygons  are  mutually  equiangular. 

Second.     In  A  GHI  and  G' H 


GH 
G'H' 


HI 
H'I' 


In  A  GIJ  and  eW,      --  =          = 


G'l'      I'J' 


In  A  GJK  and  6?' J'K',  —  = 
G'J' 


GH 


HI 


IJ 


G'J1 
KG 

~  K'G> 

JK 


KG 
K'G' 


That  is,  the  homologous  sides  are  proportional. 
.•.  The  polygons  are  similar 


(Ax.  1). 
(301). 

Q.E.D. 


Ex.   If  two  parallelograms  are  similar,  and  a  diagonal  in  each  is  drawn, 
are  the  resulting  triangles  necessarily  similar  in  pairs? 


166 


BOOK   111.     PLANE   GEOMETRY 


PROPOSITION  XXXII.     THEOREM 

320.    If  through  a  fixed  point  within  a  circle  two  chords  are 
drawn,  the  product  of  the  segments  of  one 
equals  the  product  of  the  segments  of  the 
other. 

Given  :  Point  O  in  circle  C  ;  chords  AB 
and  BS  intersecting  at  O.  (Review  the 
note,  p.  149.) 

To  Prove  :  AO  •  OB  =  RO  -  os. 
Proof  :  Draw  AS  and  RB. 
In  A  AOS  and  BOB,          Z  s  =  /-  B 
And  Z^=ZE 

.•.  these  A  are  similar 
.-.  AO  :  BO  =  OS  :  OB 
.'.  AO-  OB  =  EO  •  OS 


(239). 
' 


(303). 
(313). 
(280).      Q.E.D. 

321.  COROLLARY.     The  product  of  the  segments  of  any  chord 
drawn  through  a  fixed  point  within  a  circle  is  constant  for  all 
chords  through  this  point. 

322.  Direct  proportion  and  reciprocal  (or  inverse)  proportion. 

Illustrations.  I.  If  a  man  earns  $4|  each  day,  in  8  days  he  earns 
$36.  In  12  days  he  earns  $54.  Hence,  8  da.  :  12  da.  =  $36:  $54  is 
a  proportion  in  which  the  antecedents  belong  to  the  same  condition  or 
circumstance,  and  the  consequents  belong  to  some  other  condition  or 
circumstance.  This  is  called  a  direct  proportion. 

II.  If  one  man  can  build  a  certain  wall  in  120  days,  8  men  can  build 
it  in  15  days  ;  or  12  men  in  10  days.  Hence,  8  men  :  12  men  =  10  da.  : 
15  da.  is  a  proportion  in  which  the  means  belong  to  the  same  condition 
or  circumstance,  and  the  extremes  belong  to  some  other  condition  or  cir- 
cumstance. This  is  called  a  reciprocal  (or  inverse)  proportion. 

Definitions.  A  direct  proportion  is  a  proportion  in  which 
the  antecedents  belong  to  the  same  circumstance  or  figure, 
and  the  consequents  belong  to  some  other  circumstance  or 
figure.  Thus,  the  ordinary  proportions  derived  from  similar 
figures  are  direct  proportions. 


CIRCLES 


167 


A  reciprocal   (or  inverse)  proportion  is   a  proportion  in 
which  the  means  belong  to  the  same  cir- 
cumstance or  figure,  and  the  extremes 
belong   to  some   other   circumstance  or 
figure. 

Thus,  in  the  adjoining  figure,  a  •  b  =  x  >y  (320). 
.*.  a  :  x  —  y  :  b  (281).  This  is  a  reciprocal  pro- 
portion because  the  means  are  parts  of  one  chord, 
and  the  extremes  are  parts  of  the  other  chord. 

323.  THEOREM.  If  through  a  fixed  point  within  a  circle  two 
chords  are  drawn,  their  four  segments  are  reciprocally  (or 
inversely)  proportional. 

Proof:  [Identical  with  320;  omitting  the  last  step.] 


PROPOSITION  XXXIII.     THEOREM 

324.    If  from  a  fixed  point  without  a  circle  a  secant  and  a 
tangent  are  drawn,  the  product  of 
the  whole  secant  and  the  external 
segment  equals  the  square  of  the 
tangent. 

Given :   O  C ;  secant  PAB ;  tan- 
gent PT. 

To  Prove :   PB  •  PA  =  PT2. 

Proof:  Draw  AT  and  BT. 

In  A  PAT  and  PBT,         Z  P  =  Z  P 

Z.PTA  is  measured  by  ^  arc  AT 
Z  B  is  measured  by  ^  arc  ^.T 


Therefore 
Hence 


A  PAT  is  similar  to  A  PBT 
PB  :  PT  =  PT  :  PA 


(Iden.). 
(241). 
(236). 
(237). 
(303). 
(313). 
(280). 

Q.E.D. 


168  BOOK  III.     PLANE   GEOMETKY 

325.  COROLLARY.  If  from  a  fixed  point  without  a  circle  any 
secant  is  drawn,  the  product  of  the  secant  and  its  external 
segment  is  constant  for  all  secants. 

Proof:  Any  secant  x  ext.  seg.  =  (tan.)2  =  constant. 


326.  COROLLARY.    If  from  a  fixed  point  without  a  circle  two 
secants  are  drawn,  these  secants  and  their  external  segments 
are  reciprocally  (or  inversely)  proportional. 

Proof :  PB  -  PA  =  PY  -  PX  (325). 

.-.  PB  :  PY  =  PX  :  PA  (281). 

•Q.E.D. 

327.  THEOREM.    If  from  a  fixed  point  without  a   circle   a 
secant  and  a   tangent   are   drawn,  the   tangent   is   a   mean 
proportional  between  the  secant  and  its  external  segment. 

Proof :    [Identical  with  proof  of  324  ;    omitting  the  last 
step.] 


Ex.  1.   If  PA  =  3  in.,  and  PB  =  12  in.,  find  the  length  of  PT. 
Ex.  2.    If  PB  =  21  in.,  PY  =  15  in.,  and  PA  =  5  in.,  find  PX. 

Ex.  3.  Two  altitudes  of  a  triangle  are  reciprocally  proportional  to  the 
bases  to  which  they  are  drawn. 

Ex.  4.  If  AD  and  BE  are  two  altitudes  of  a  triangle,  and  DE 
is  drawn,  the  triangle  ABC  is  similar  to  the  triangle  CED.  [Use 
306.] 

Ex.  5.  The  four  segments  of  the  diagonals  of  a  trapezoid  are  propor- 
tional. 


CIRCLES 


169 


PROPOSITION  XXXIV.     THEOREM 

328.  In  any  triangle  the  product  of  two 
sides  is  equal  to  the  diameter  of  the  cir- 
cumscribed circle  multiplied  by  the  alti- 
tude upon  the  third  side. 

Given :  A  ABC  ;  circumscribed  Q  o  ; 
altitude  BK. 

To  Prove :  a  •  c  =  d  •  h. 

Proof :  Draw  chord  CD.     Z  BCD  =  rt.  Z 

In  rt.  A  ABK  and  BCD,  Z  A  =  Z  D 

.•.  these  A  are  similar 


(240). 
(239). 
(304). 
(313). 
(280).  Q.E.D. 


PROPOSITION  XXXV.     THEOREM 

329.  In  any  triangle  the  product  of  two  sides  is 
square  of  the  bisector  of  their  included 
angle,    plus   the   product    of    the    seg- 
ments of  the  third  side  formed  by  the 
bisector. 

Given:  A  ABC,  CO  bisector  of  Z  ACS. 
To  Prove :  a  -  b  =  t2  -f-  n  •  r. 

Proof :  Circumscribe  a  O  about  the  A  ABC. 
Produce  CO  to  meet  O  at  D.     Draw  BD. 
In  A  AGO  and  BCD,  Z  AGO  =  Z  BCD 

Also  Z  A  =  Z  D 

.-.  &ACO  and  BCD  are  similar 


Now 


Substituting, 


AB  and  CD  are  chords 


a  •  b  =  t2  +  n  •  r 


equal  to  the 

...........  -    C 


/o    A 

I     /  I 


(Ax. 


(Hyp.). 
(239). 
(303). 
(313). 
(280). 
(Const.). 
(320). 
6).     Q.E.D. 


BOBBINS'S   NEW    PLANE    GEOM.  —  12 


170 


BOOK   III.     PLANE   GEOMETRY 


330.    The  projection  of  a  point  upon  a  line  is  the  foot  of 
the  perpendicular  from  the  point  to  the  line. 
Thus,  the  projection  of  P  is  J. 


B  S            L 

^1  /\   \ 

I         i i          /         = ;J_ 

J        C           D  R           T         NM 


The  projection  of  a  definite  line  upon  an  indefinite  line  is 
the  part  of  the  indefinite  line  between  the  feet  of  the  two 
perpendiculars  to  it,  from  the  extremities  of  the  definite  line. 

The  projection  of  AB  is  CD  ;  of  RS  is  ET  ;  of  LM  is  NM. 


PROPOSITION  XXXVI.     THEOREM 

331.  If  in  a  right   triangle  a  perpendicular  is  drawn  from 
the  vertex  of  the  right  angle  upon  the  hypotenuse  : 

I.  The  triangles  formed  are  similar  to  the  given  triangle 
and  similar  to  each  other. 

n.   The  perpendicular  is  a  mean  proportional  between  the 
segments  of  the  hypotenuse. 

r* 

Given :  Rt.  A  ABC ;  CP  _L  to  AB  from  C. 
To  Prove :  A 

I.    A  APC,  ABC,  and  BPC  similar.  /_ 

II.  AP  :  CP  =  CP  :  PB.  A 


Proof :  I.    In  rt.  A  APC  and  ABC,  ^-A  =  /.A  (Iden.). 

.  •.  A  APC  is  similar  to  A  ABC  (304). 

In  rt.  A  BPC  and  ABC,  ^  B  =  Z  B  (Iden.). 

.  •.  A  BPC  is  similar  to  A  ABC  (304). 

Therefore  A  APC,  ABC,  and  BPC  are  all  similar  (307). 

II.    In  the  A  APC  and  BPC,  AP  :  CP  =  CP  :  PB  (313). 

Q.E.D. 


PROJECTIONS 


171 


332.   COROLLARY.    If  from  any  point  in  a  circumference  a 
perpendicular  is  drawn  to  a  diameter,  p 

it  is  a  mean  proportional  between  the 
segments  of  the  diameter. 
Given:  (?).     To  Prove:  (?). 
Proof :  Draw  chords  AP  and  BP. 

A  APB  is  a  rt.  A 
.-.  AD  :  PD  =  PD  :DB 


(240). 
(331,  II). 

Q.E.D. 


PROPOSITION  XXXVII.     THEOREM 


333.  The  square  of  a  leg  of  a  right  triangle  is  equal  to  the 
product  of  the  hypotenuse  and  the  projection  of  this  leg  upon 
the  hypotenuse. 

Given  :  Rt.  A  ABC  ;  AC  and  BC  the 

legs. 

I.    AC?  =  AB  -  AP. 

AB-  BP. 


To  Prove 


II. 


Proof:  I.    The  rt.  A  ABC  and  APC  are  similar 
.'.  AB  :_AC  =  AC  :  AP 
•  *.  AC^  =  AB  •  AP 

The  rt.  A  ABC  and  BCP  are  similar 
.'.  AB  :  BC=  BC  :  BP 

2J(?2  __  ^jj  .  -gp 


II. 


(331, I). 
(313). 
(280). 

(313)! 
(280). 

Q.E.D. 


Ex.  1.    If,  in  331,  AP  =  3,  PB  =  27,  find  CP. 

Ex.  2.   If,  in  333,  AP  =  4,  PB  =  21,  find  A  C  and  BC. 

Ex.  3.   If,  in  333,  AB  =  2Q,AC  =  10,  find  AP,  BP,  CP,  and  BC. 

Ex.  4.   If,  in  331,  AP  =  10  and  CP  =  20,  find  BP. 

Ex.  5.   If,  in  333,  AB  =  45  and  AC  =  15,  find  AP,  BP,  CP,  and  BC. 

Ex.  6.   If,  in  333,  AP  =  9  and  BP  =  16,  find  AC,  PC,  and  BC. 

Ex.  7.    A  stone  arch  in  the  shape  of  the  arc  of  a  circle  is  4  ft.  high. 
The  chord  of  half  the  arch  is  1.0  ft.     Find  the  diameter. 


172 


BOOK   111.     PLANE   GEOMETRY 


PROPOSITION  XXXVIII.     THEOREM 

334.  The  sum  of  the  squares  of  the 
legs  of  a  right  triangle  is  equal  to  the 
square  of  the  hypotenuse. 


Given :  Rt.  A  ABC. 

To  Prove :  a2  +  V*  =  c2. 

Proof :  Draw  CP  _1_  to  the  hypotenuse  AB. 

Denote  AP  by  p  and  PB  by  p'. 

Now  b2  =  c  -  p 

Also  a2  =  c  •  p' 

Adding, 


a 


=  c-p  +  c  .pf 
=  c(p  +pf) 

=  C  •  C  =  C2. 


(333). 

(?). 

(Ax.  2). 

Q.E.D. 


335.  COROLLARY.  The  square  of  either  leg  of  a  right  triangle 
is  equal  to  the  square  of  the  hypotenuse  minus  the  square  of 
the  other  leg. 

That  is,  a2  =  c2  -  b*  ;  and  b2  =  c*  -  a2  (Ax.  2). 


Ex.  1.   If  AC  =  28  and  EC  =  45,  find  AB. 
Ex.  2.   If  AC  =  21  and  AB  =  29,  find  EC. 

Ex.  3.    The  square  of  the  altitude  of  an  equilateral  triangle  equals 
three  fourths  the  square  of  a  side. 

Ex.  4.  In  any  triangle  the  difference  of  the 
squares  of  two  sides  is  equal  to  the  'difference  of 
the  squares  of  their  projections  on  the  third  side. 

[ZS2  =  (?)  ;  BC2  =  (?).     Subtract,  etc.] 

Ex.  6.  If  the  altitudes  of  triangle  ABC  meet 
at  0,.AB*  -  AC2  =  B02-  CO2.  A~~ 

[Consult  Ex.  4  and  substitute.] 

Ex.  6.  If  lines  are  drawn  from  any  point  in  a  circle 
to  the  four  vertices  of  an  inscribed  square,  the  sum  of 
the  squares  of  these  four  lines  is  equal  to  twice  the 
square  of  the  diameter. 

Proof:  &APC,  DPB  are  vt,  A;  etc. 


SQUARES  OF   SIDES   OF  TRIANGLES  173 

PROPOSITION  XXXIX.    THEOREM 

336.   In  an  obtuse  triangle  the  square  of  the  side  opposite 
the  obtuse  angle  is  equal  to  the  sum  of  the  squares  of  the  other 
two  sides  plus  twice  the  product  of  one 
of  these  two  sides  and  the  projection  of 
the  other  side  upon  that  one. 

Given :  Obtuse  A  ABC ;  etc.  ^- — - 


C        M 
To  Prove :  c2  =  a2  +  b2  +  2  bp. 

Proof :  In  right  A  CBM,  h2+p2  =  a2  (334). 

In  right  A  ABM,  (?  =  h*  +  (p  +  b)2  (334). 

=  h2  +  p2  +  b2  +  2  bp 

=  ~a?~^ +  b2  +  2bp       (Ax.  6). 

Q.E.D. 

PROPOSITION  XL.    THEOREM 

337.  In  any  triangle  the  square  of  the  side  opposite  an  acute 
angle  is  equal  to  the  sum  of  the  squares  of  the  other  two  sides 
minus  twice  the  product  of  one  of  these  two  sides  and  the  pro- 
jection of  the  other  side  upon  that  one. 

Given:  (?). 

To  Prove :  c?  =  a2  +  b2  —  2  bp.  ^  t>-t>        I  P 

Proof :  In  one  rt.  A,  h2  +  p2  =  a2  (334) . 


In  the  other  rt.  A,  c2  =  W  +  (b-p)*  (334). 


(Ax.  6). 

Q.E.D. 
NOTE.     This  theorem  is  equally  true  in  the  case  of  an  obtuse  triangle. 


Ex.  If  lines  are  drawn  from  any  external  point 
to  the  vertices  of  a  rectangle  ABCD,  the  sum  of  the 
squares  of  two  of  them  which  are  drawn  to  a  pair 
of  opposite  vertices  is  equal  to  the  sum  of  the 
squares  of  the  other  two. 

To  Prove :  PA2  +  PC2  =  PB2  +  P&. 


174  BOOK  III.     PLANE   GEOMETRY 


PROPOSITION  XLI.     THEOREM 

338.  I.  The  sum  of  the  squares  of  two  sides  of  a  triangle 
is  equal  to  twice  the  square  of  hah"  the  third  side  increased  by 
twice  the  square  of  the  median  upon  that  side. 

II.  The  difference  of  the  squares  of  two  sides  of  a  triangle 
is  equal  to  twice  the  product  of  the  third  side  by  the  projection 
of  the  median  upon  that  side. 

Given :    A  ABC ;    median  =  m ;    its 
projection  =  p ;  and  side  b  >  side  a. 

To  Prove : 


Proof :  In  A  ARC  and  BRC,  AR  =  BE  (Hyp.). 

And                                  CR  =  CR  (Iden.). 

Also                              AC  >  BC  (Hyp.). 

.-.  Z  ARC  >  Z  BRC  (92). 
That  is,         Z.ARC  is  obtuse  and  Z  BRC  is  acute. 

52=    (<^)2+    w2  +  ^  (336). 

a?=   (^e)2+    m*-cp  (337). 

I.    Adding,            62+ a2  =  2(-|V)2+2w2  (Ax.  2). 

II.    Subtracting,     b2  —  a2  =                           2cp  (Ax.  2). 

Q.E.D. 

339.  Formulas.  If  the  vertices  of  a  triangle  are  denoted 
by  A,  -B,  C,  the  lengths  of  the  sides  opposite  are  denoted  by 
a,  5,  0,  respectively  ;  the  altitude  upon  these  sides  by  A0,  hb  hc, 
respectively ;  the  bisectors  of  the  angles  by  ta,  tb,  tc,  respec- 
tively ;  the  medians  by  ma,  mb,  m^  respectively ;  the  seg- 
ments of  the  sides  formed  by  the  bisectors  of  the  opposite 
angles  by  na  and  r0,  nb  and  rb,  nc  and  rc\  and  the  projections 
as  follows:  the  projection  of  side  a  upon  side  5,  by  apb\  of 
side  a  upon  side  <?,  by  apc ;  of  side  b  upon  side  e,  by  bpc ;  etc. 


FORMULAS  175 

It  is  assumed  that  #,  6,  c  are  known.    The  following  values 
of  the  various  lines  in  a  triangle  are  obtained  by  solving  the 
equations  already  established. 
I.   Projections. 

/>2         rt2         7>2  r2        rfi        W 

1.  If  /.  C  is  obtuse,  a^6  =  C-—jL-± ;  6pa  =  £ |_A;  etc. 

«2    i     ;.2         ,.2  ^2    i    1,2          ~2 

2.  If  Z  C  is  acute,    apb  =  - +^~-;  ^  =  -  +^  g~     ;  etc. 
IT.   Altitudes.  hb  =    Va2  -  apb2 ;  A0  =  V&2  -  6 p02 ;  etc. 


III.    Medians.   mc  =  iV2(«'2  +  b2)  -  c2;  m0  =  ^V2(&2  +  c2)  -  a2;  etc. 


IV.   Bisectors.     tc  =  Vab  -  ncr* ;  ta  =  V6c  -  nara* ;  f6  =  Vac  -  nbrb* 


V.   Diameter  of  circumscribed  circle  =  — ;    =  — ;    =  —  • 

hb          hc          ha 

VI.   Largest  Angle.     Denote  largest  angle  by  C. 

1.  If  c2  =  a2  +  62,  Z  C  is  right  (334). 

2.  If  c2  =  a2  f  b2  plus  something,  Z  C  is  obtuse  (336). 

3.  If  c2  =  a2  +  62  minus  something,  Z.  C  is  acute  (337). 


Ex.  1.  If  the  sides  of  a  triangle  are  a  =  7,  b  =  10,  c  =  12,  find  the 
nature  of  Z  C. 

Ex.  2.   In  the  same  triangle  find  raa.     Find  mb.     Find  mc. 

Ex.  3.   In  the  same  triangle  find  apb.     Find  bpa.     Find  ajt>c.     Find  frj9c. 
Find  cpa.     Find  «./>&. 
1  Ex.  4.    Find  /*„.     Find  hb.     Find  ^c. 

Ex.  5.    Find  the  diameter  of  the  circumscribed  circle. 

Ex.  6.    Find  na  and  ra:    Find  nb  and  rb.     Find  nc  and  rc. 

Ex.  7.    Find  *0.     Find  tb.     Find  <c. 

CONCERNING   ORIGINALS 

340.  First  determine  from  the  nature  of  each  numerical 
exercise  upon  which  theorem  it  depends,  and  then  apply 
that  theorem. 

*  The  segments  n  and  r  can  be  found  by  297  ;  nb :  rb  =  c  :  a,  etc. 


176  BOOK   III.     PLANE   GEOMETRY 

ORIGINAL  EXERCISES    (NUMERICAL) 

1.  The  legs  of  a  right  triangle  are  12  and  16  inches.    Find  the  hypote- 
nuse. 

2.  The  side  of  a  square  is  6  feet.     What  is  the  diagonal? 

3.  The  base  of  an  isosceles  triangle  is  16  and  the  altitude  is  15.    Find 
the  equal  sides. 

4.  The  tangent  to  a  circle  from  a  point  is  12  inches  and  the  radius 
of  the  circle  is  5  inches.     Find  the  length  of  the  line  joining  the  point  to 
the  center. 

5.  In  a  circle  whose  radius  is  13  inches,  what  is  the  length  of  a  chord 
5  inches  from  the  center? 

6.  The  length  of  a  chord  is  2  feet  and  its  distance  from  the  center 
is  35  inches.     Find  the  radius  of  the  circle. 

7.  The  hypotenuse  of  a  right  triangle  is  2  feet  2  inches,  and  one  leg 
is  10  inches.     Find  the  other. 

8.  The  base  of  an  isosceles  triangle  is  90  inches  and  the  equal  sides 
are  each  53  inches.     Find  the  altitude. 

9.  The  radius  of  a  circle  is  4  feet  7  inches.     Find  the  length  of  the 
tangent  drawn  from  a  point  6  feet  1  inch  from  the  center. 

10.  How  long  is  a  chord  21  yards  from  the  center  of  a  circle  whose 
radius  is  35  yards  ? 

11.  Each  side  of  an  equilateral  triangle  is  4  feet.     Find  the  altitude. 

12.  The  altitude  of  an  equilateral  triangle  is  8  feet.     Find  the  side. 

13.  Each  side  of  an  isosceles  right  triangle  is  a.    Find  the  hypotenuse. 

14.  If  the  length  of  the  common  chord  of  two  intersecting  circles  is 
16,  and  their  radii  are  10  and  17,  what  is  the  distance  between  their 
centers? 

15.  The  diagonal  of  a  rectangle  is  82  and  one  side  is  80.     Find  the 
other. 

16.  The  length  of  a  tangent  to  a  circle  whose  diameter  is  20,  from  an 
external  point,  is  26.     What  is  the  distance  from  this  point  to  the  center  ? 

17.  The  diagonal  of  a  square  is  10.     Find  each  side. 

18.  Find  the  length  of  a  chord  2  feet  from  the  center  of  a  circle 
whose  diameter  is  5  feet. 

19.  A  flagpole  was  broken  16  feet  from  the  ground,  and  the  top  struck 
the  ground  63  feet  from  the  foot  of  the  pole.     How  long  was  the  pole  ? 


ORIGINAL  EXERCISES 


177 


20.  The  top  of  a  ladder  17  feet  long  reaches  a  point  on  a  wall  15  feet 
from  the  ground.     How  far  is  the  lower  end  of  the  ladder  from  the  wall? 

21.  A  chord  2  feet  long  is  5  inches  from  the  center  of  a  circle.     How 
far  from  the  center  is  a  chord  10  inches  long?     [Find  the  radius.] 

22.  The  diameters  of  two  concentric  circles  are  1  foot  10  inches  and 
10  feet  2  inches.     Find  the  length  of  a  chord  of  the  larger  which  is  tan- 
gent to  the  less. 

23.  The  lower  ends  of  a  post  and  a  flagpole  are  42  feet  apart;  the 
post  is  8  feet  high  and  the  pole,  48  feet.     How  far  is  it  from  the  top  of 
one  to  the  top  of  the  other? 

24.  The  radii  of  two  circles  are  8  inches  and  17  inches,  and  their  cen- 
ters are  41  inches  apart.     Find  the  lengths  of  their  common  external 
tangents ;  of  their  common  internal  tangents. 

25.  A  ladder  65  feet  long  stands  in  a  street.     If  it  inclines  toward  one 
side,  it  will  touch  a  house  at  a  point  16  feet  above  the  pavement ;  if  to 
the  other  side,  it  will  touch  a  house  at  a  point  56  feet  above  the  pave- 
ment.    How  wide  is  the  street? 

26.  Two  parallel  chords  of  a  circle  on  opposite  sides  of  the  center  are 
4  feet,  and  40  inches  long,  respectively,  and  the  distance  between  them  is 
22  inches.     Find  the  radius  of  the  circle. 

[Draw  the  radii  to  ends  of  chords ;   these  =  hypotenuses  =  R ;   the 
distances  from  the  center  =  x  and  22  —  #.] 

27.  The  legs  of  an  isosceles  trapezoid  are  each  2  ft.  1  in.  long,  and  one 
of  the  bases  is  3  ft.  4  in.  longer  than  the  other.     Find  the  altitude. 

28.  One  of  the  non-parallel  sides  of  a  trapezoid  is  perpendicular  to 
both  bases,  and  is  63  feet  long ;  the  bases  are  41  feet  and  25  feet  long. 
Find  the  length  of  the  remaining  side. 

29.  If  a  =  10,  h   =    6,  find  p,  c,  p',  b. 

30.  If  h  =    8,  p'  =    4,  find  b,  c,  p,  a. 

31.  If  a  =  10,  p'  =  15,  find  p,  c,  h,  b. 

32.  If  a  =    9,  b  =  12,  find  c,  p,p',  h.  . 

33.  Ifp=    3,  p'  =  12,  find  h,  a,  b. 

34.  The  line  joining  the  midpoint  of  a  chord  to  the 
midpoint  of  its  arc  is  5  inches.     If  the  chord  is  2  feet 
long,  what  is  the  diameter? 

[A  ACE  is  rt.  A  (?).     .-.  AC2  =  CE  .  CD  (?).] 

35.  If  the  chord  of  an  arc  is  60  and  the  chord  of  its 
half  is  34,  what  is  the  diameter? 


178  BOOK  III.     PLANE   GEOMETRY 

36.  The  line  joining  the  midpoint  of  a  chord  to  the  midpoint  of  its  arc 
is  6  inches.     The  chord  of  half  this  arc  is  18  inches.     Find  the  diameter. 
Find  the  length  of  the  original  chord. 

37.  To  a  circle  whose  radius  is  10  inches,  two  tangents  each  2  feet 
long  are  drawn  from  a  point.     Find  the  length  of  the  chord  joining 
their  points  of  contact. 

38.  The  sides  of  a  triangle  are  6,  9,  11.     Find  the  segments  of  the 
shortest  side  made  by  the  bisector  of  the  opposite  angle. 

39.  Find  the  segments  of  the  longest  side  made  by  the  bisector  of  the 
largest  angle  in  Ex.  38. 

40.  The  sides  of  a  triangle  are  5,  9,  12.     Find  the  segments  of  the 
shortest  side  made  by  the  bisector  of  the  opposite  exterior  angle ;  also 
of  the  medium  side  made  by  the  bisector  of  its  opposite  exterior  angle. 

41.  In  the  figure  of  295,  if  AC  =  3,  CE  =  5,  EG  =  8,  BD  =  4,  find 
DF  and  FH. 

42.  If  the  sides  of  a  triangle  are  6,  8,  12  and  the  shortest  side  of  a 
similar  triangle  is  15,  find  its  other  sides. 

43.  If  the  homologous  altitudes  of  two  similar  triangles  are  9  and  15 
and  the  base  of  the  former  is  21,  what  is  the  base  of  the  latter? 

44.  In  the  figure  of  315,  AE  =  4,  EF  =  6,  FB  =  9,  GH  =  15.     Find 
CG  and  CD. 

45.  The  sides  of  a  pentagon  are  5,  6,  8,  9,  18,  and  the  longest  side  of 
a  similar  pentagon  is  78.     Find  the  other  sides. 

46.  A  pair  of  homologous  sides  of  two  similar  polygons  are  9  and  16. 
If  the  perimeter  of  the  first  is  117,  what  is  the  perimeter  of  the  second? 

47.  The  perimeters  of  two  similar  polygons  are  72  and  120.     The 
shortest  side  of  the  former  is  4.     What  is  the  shortest  side  of  the  latter? 

48.  Two  similar  triangles  have  homologous  bases  20  and  48.     If  the 
altitude  of  the  latter  is  36,  find  the  altitude  of  the  former. 

49.  The  segments  of  a  chord,  made  by  a  second  chord,  are  4  and  27. 
One  segment  of  the  second  chord  is  6.    Find  the  other. 

50.  One  of  two  intersecting  chords  is  19  inches  long  and  the  segments 
of  the  other  are  5  inches  and  12  inches.     Find  the  segments  of  the  first 
chord. 

51.  Two  secants  are  drawn  to  a  circle  from  a  point ;  their  lengths  are 
15  inches  and  10|  inches.     The  external  segment  of  the  latter  is  10  inches. 
Find  the  external  segment  of  the  former. 


ORIGINAL  EXERCISES  179 

52.  The  tangent  to  a  circle  is  1  foot  long  and  the  secant  from  the 
same  point  is  1  foot  6  inches.     Find  the  chord  part  of  the  secant. 

53.  The  internal  segment  of  a  secant  25  inches  long  is  16  inches. 
Find  the  tangent  from  the  same  point  to  the  same  circle. 

54.  Two  secants   to   a   circle  from  a  point  are  1|  feet  and  2  feet 
long ;  the  tangent  from  the  same  point  is  12  inches.     Find  the  external 
segments  of  the  two  secants. 

55.  If  the  sides  of  a  triangle  are  5,  6,  8,  is  the  angle  opposite  8  right, 
acute,  or  obtuse  ?  if  the  sides  are  8,  7,  4  ? 

56.  If  the  sides  of  a  triangle  are  8,  9,  12,  is  the  largest  angle  right, 
acute,  or  obtuse?  if  the  sides  are  13,  7,  11  ? 

57.  The  sides  of  a  triangle  are  x,  y,  z.     If  z  is  the  greatest  side,  when 
will  the  angle  opposite  be  right?  obtuse?  acute? 

58.  The  sides  of  a  triangle  are  6,  8,  9.     Find  the  length  of  the  projec- 
tion of  side  6  upon  side  8 ;  of  side  8  upon  side  9 ;  of  side  9  upon  side  6. 

59.  The  sides  of  a  triangle  are  5,  6,  9.     Find  the  length  of  the  pro- 
jection of  side  6  upon  side  5 ;  of  side  9  upon  side  6. 

60.  Find  the  three  altitudes  in  a  triangle  with  sides  9,  10,  17. 

61.  Find  the  three  altitudes  in  a  triangle  with  sides  11,  13,  20. 

62.  Find  the  diameter  of  a  circle  circumscribed  about  a  triangle  with 
sides  17,  25,  26. 

63.  Find  the  length  of  the  bisector  of  the  least  angle  of  a  triangle 
with  sides  7,  15,  20;  also  of  the  largest  angle. 

64.  Find  the  length  of  the  bisector  of  the  largest  angle  of  a  triangle 
with  sides  12,  32,  33 ;  also  of  the  other  angles. 

65.  Find  the  three  medians  in  a  triangle  with  sides  4,  7,  9. 

66.  Find  the  product  of  the  segments  of  every  chord  drawn  through 
a  point  4  units  from  the  center  of  a  circle  whose  radius  is  10  units. 

67.  The  bases  of  a  trapezoid  are  12  and  20,  and  the  altitude  is  8.     The 
other  sides  are  produced  to  meet.     Find  the  altitude  of  the  larger  tri- 
angle formed. 

68.  The  shadow  of  a  yardstick  perpendicular  to  the  ground  is  4|  feet. 
Find  the  height  of  a  tree  whose  shadow  at  the  same  time  is  100  yards. 

69.  There  are  two  belt-wheels  3  feet  8  inches  and  1  foot  2  inches  in 
diameter,  respectively.     Their  centers  are  9  feet  5  inches  apart.     Find 
the  length  of  the  belt  suspended  between  the  wheels  if  the  belt  does  not 
cross  itself ;  also  the  length  of  the  belt  if  it  does  cross. 


180 


BOOK   III.     PLANE   GEOMETRY 


SUMMARY 

341.  Triangles  are  proved  similar  b}^  showing  that  they 
have : 

(1)  Two  angles  of  one  equal  to  two  angles  of  the  other. 

(2)  An  acute  angle  of  one  equal  to  an  acute  angle  of  the  other.     [In 
right  triangles.] 

(3)  Homologous  sides  proportional. 

(4)  An  angle  of  one  equal  to  an  angle  of  the  other  and  the  including 
sides  proportional. 

(5)  Their  sides  respectively  parallel  or  perpendicular. 

Four  lines  are  proved  proportional  by  showing  that  they  are : 

(1)  Homologous  sides  of  similar  triangles. 

(2)  Homologous  sides  of  similar  polygons. 

(3)  Homologous  lines  of  similar  figures. 

The  product  of  two  lines  is  proved  equal  to  the  product  of  two 
others,  by  proving  these  four  lines  proportional  and  making 
the  product  of  the  extremes  equal  to  the  product  of  the  means. 

One  line  is  proved  a  mean  proportional  between  two  others 
by  proving  that  two  triangles  containing  this  line  in  common 
are  similar,  and  obtaining  the  proportion  from  their  sides. 

In  cases  dealing  with  the  square  of  a  line,  one  uses  : 

(1)  Similar  triangles  having  this  line  in  common,  or, 

(2)  A  right  triangle  containing  this  line  as  a  part. 

ORIGINAL  EXERCISES   (THEOREMS) 

1.  In   any  right  triangle    the    product    of  the 
hypotenuse  and  the  altitude  upon  it  is  equal  to  the 
product  of  the  legs. 

2.  If  two  circles  intersect  at  A  and  J5,  and  AC 
and  AD  are  drawn,  each  a  tangent  to  one  circle  and 
a  chord  of  the  other,  the  common  chord  AB  is  a 
mean  proportional  between  BC  and  BD. 

3.  If  AB  is  a  diameter  and  BC  a,  tangent,  and 
A  C  meets  the  circumference  at  D,  the  diameter  is  a 
mean  proportional  between  AC  and  AD. 


ORIGINAL   EXERCISES 


181 


4.  If  two  circles  are  tangent  externally,  the  chords 
formed  by  a  straight  line  drawn  through  their  point  of 
contact  have  the  same  ratio  as  the  diameters  of  the 
circles. 

6.  If  a  tangent  is  drawn  from  one  extremity  of  a 
diameter,  meeting  secants  from  the  other  extremity, 
these  secants  and  their  internal  segments  are  recip- 
rocally proportional. 

To  Prove:  AC  :  AD  =  AS  :  AR. 

Proof :  Draw  RS.    In&ARS  and  A  CD,  ^A=ZAai\ 
(Explain.)     Etc. 

6.  If  AB  is  a  chord  and  CE,  another  chord,  drawn 
from  C,  the  midpoint  of  arc  AB,  meeting  chord  AB  at 
D,  A  C  is  a  mean  proportional  between  CD  and  CE. 

Prove  the  above  theorem  and  deduce  that,  CE  •  CD 
is  constant  for  all  positions  of  the  point  E  on  arc  AEB 

7.  If  chord  AD  is  drawn  from  vertex  A   of  in- 
scribed isosceles  triangle  ABC,  cutting  BC  at  E,  AB 
is  a  mean  proportional  between  AD' and  AE. 

Prove  the  above  theorem  and  deduce  that,  AD  •  AE 
is  constant  for  all  positions  of  the  point  D  on  arc 
BDC. 

8.  If  a  square  is  inscribed  in  a  right  triangle 
so  that  one  vertex  is  on  each  leg  of  the  triangle  and 
the  other  two  vertices  on  the  hypotenuse,  the  side 
of  the  square  is  a  mean  proportional  between  the 
other  segments  of  the  hypotenuse. 

To  Prove :  AD  :  DE  =  DE :  EB. 
First  prove  A  ADG  and  BEF  similar. 

9.  If  the  sides  of  two  unequal  triangles  are  respectively  parallel,  the 
lines  joining  homologous  vertices  meet  in  a  point.     (These  lines  to  be 
produced  if  necessary.) 

10.  AB  is  any  chord;  AC  is  a  tangent  and  CDE  is  a  secant  parallel 
to  AB  cutting  the  circle  at  D  and  E.     Prove  that  A  C  :  AE  =  DC  :  BE. 

11.  Prove  theorem  of  320,  by  drawing  two  other  auxiliary  lines. 

12.  Prove  theorem  of  316  if  point  0  is  between  the  parallels. 

13.  Prove  theorem  of  327  by  drawing  auxiliary  lines  A  Y  and  BX. 


182 


BOOK   III.     PLANE   GEOMETRY 


14.  If  one  leg  of  a  right  triangle  is  double 
the  other,  its  projection  upon  the  hypotenuse 
is  four  times  the  projection  of  the  other. 


Proof: 


(2  a)2  =  cp-,  a?=cp< 

;      p'=^L.         (AX.  3). 


C  C 

15.  If  the  bisector  of  an  angle  of  a  triangle  bisects  the  opposite  side, 
the  triangle  is  isosceles. 

16.  The  tangents  to  two  intersecting  circles  from  any  point  in  their 
common  chord  produced  are  equal.     [Use  324.] 

17.  If  two  circles  intersect,  their  common  chord,  produced,  bisects 
their  common  tangents. 

18.  If  AB  and  A  C  are  tangents  to  a  circle 
from   A  ;    CD  is   perpendicular  to   diameter 
BOX  from  C;  then  AB  •  CD  =  BD  •  BO. 

19.  If   the   altitude  of   an  equilateral  tri- 
angle is  h,  find  the  side.  X" 

20.  If  one  side  of  a  triangle   is  divided  by  a  point  into   segments 
which  are  proportional  to  the  other  sides,  a  line  from  this  point  to  the 
opposite  angle  bisects  that  angle. 

To  Prove :  Z  n  =  Z  m  in  figure  of  297. 
Proof:  Produce  CB  to  P,  making  BP  -  AB;  draw  AP;  etc. 

21.  State  and  prove  the  converse  of  298. 

22.  Two  rhombuses  are  similar  if  an  angle  of  one  is  equal  to  an  angle 
of  the  other. 

23.  If  two  circles  are  tangent  internally  and  any  two  chords  of  the 
greater  are  drawn  from  their  point  of  contact,  they  are  divided  propor- 
tionally by  "the  less  circle. 

[Draw  diameter  to  point  of  contact  and  prove  the  right  &  similar.] 

24.  The  non-parallel  sides  of  a  trapezoid  and  the  line  joining  the  mid- 
points of  the  bases,  if  produced,  meet  at  a  point.     [Use  Ax.  3  and  316.] 

25.  The  diagonals  of  a  trapezoid  and  the  line  joining  the  midpoint  of 
the  bases  meet  at  a  point. 

26.  If  one  chord  bisects  another,  either  segment  of  the  latter  is  a  mean 
proportional  between  the  segments  of  the  other. 

27.  Two  parallelograms  are  similar  if  they  have  an  angle  of  the  one 
equal  to  an  angle  of  the  other  and  the  including  sides  proportional. 


ORIGINAL   EXERCISES 


183 


28.  Two  rectangles  are  similar  if  two  adjoining  pairs  of   homologous 
sides  are  proportional. 

29.  If  two  circles  are  tangent  externally,  the 
common  exterior  tangent  is  a  mean  proportional 
between  the  diameters. 

[Draw  chords  PA,  PC,  PB,  PD. 

Prove  Z  APE  a  rt,  Z. 

Then  prove  APD  and  BPC  straight  lines. 

Then  prove  A  ABC  and  ABD  similar.] 

30.  In  any  rhombus  the  sum  of  the  squares  of  the  diagonals  is  equal 
to  the  square  of  half  the  perimeter. 

31.  If  in  an  angle  a  series  of  parallel  lines  are  drawn  having  their 
ends  in   the  sides   of  the   angle,  their  midpoints  lie   in   one   straight 
line. 

32.  If  ABC  is  an  isosceles  triangle  and  BX  is  the  altitude  upon  AC 
(one  of  the  legs),  BC2  =  2  A C  •  CX.     [Use  337.] 

33.  In  an  isosceles  triangle  the  square  of  one  leg  is 
equal  to  the  square  of  the  line  drawn  from  the  vertex 
to  any  point  of  the  base,  plus  the  product  of  the  seg- 
ments of  the  base. 

Proof:  Circumscribe  a  O;  use  method  of  329. 

34.  If  a  line  is  drawn  in  a  trapezoid  parallel 
to  the  bases,  the  segments  between  the  diagonals 
and  the  non-parallel  sides  are  equal. 


Proof:  &  AH  I  and  ABC  are  similar  (?)  ;  &  DKJ  and  DCB  also. 


Efcc- 


35.  A  line  through  the  point  of  intersection  of  the  diagonals  of  a 
trapezoid,  and  parallel  to  the  bases,  is  bisected  by  that  point. 

36.  If  M  is  the  midpoint  of  hypotenuse  AB  of  right  triangle  ABC, 
AB2  +  BC2  +  AC*  =  8  CM2. 

37.  The  squares  of  the  legs  of  a  right  triangle  have  the  same  ratio  as 
their  projections  upon  the  hypotenuse. 


184 


BOOK   III.     PLANE   GEOMETRY 


38.  If   the  diagonals  of   a  quadrilateral  are  perpendicular   to  each 
other,  the  sum  of  the  squares  of  one  pair  of  opposite  sides  is  equal  to  the 
sum  of  the  squares  of  the  other  pair. 

39.  The  sum  of  the  squares  of  the  four  sides  of  a  parallelogram  is 
equal  to  the  sum  of  the  squares  of  the  diagonals.     [Use  338,  I.] 

40.  If  DE  is  drawn  parallel  to  the  hypotenuse  AB  of  right  triangle 
ABC,  meeting  AC  at  D  and  CB  at  E,  AE*  +  BE?  =  AB2  +  DE2. 

[Use  4  rt.  &  having  vertex  C.~\ 

41.  If  between  two  parallel   tangents  a  third 
tangent  is  drawn,  the  radius  of  the  circle  is  a  mean 
proportional   between   the   segments  of  the  third 
tangent. 

To  Prove :  BP  :  OP  =  OP :  PD. 

Proof:  A  BOD  is  a  rt  A  (?).     Etc. 

B 
42.    If   ABCD  is  a  parallelogram,  BD  a          / 

diagonal,  A  G  any  line  from  A  meeting  BD  at        / 
E,  CD  at  F,  and  EC  (produced)  at  G,  AE  is 
a  mean  proportional  between  EF  and  EG.        ^ 


Proof:  A  ABE  and  EDF  are  similar  (?)  ;  also  &  ADE  and  BEG  (?). 
Obtain  two  ratios  equal  to  BE :  ED  and  then  apply  Ax.  1. 

43.  An  interior  common  tangent  of  two  circles  divides  the  line  join- 
ing their  centers  into  segments  proportional  to  the  radii. 

44.  An  exterior  common  tangent  of  two  circles  divides  the  line  join- 
ing their  centers  (externally)  into  segments  proportional  to  the  radii. 

46.  The  common  internal  tangents  of  two  circles  and 
the  common  external  tangents  meet  on  the  line  deter- 
mined by  the  centers  of  the  circles. 

46.  If  from  the  midpoint  P,  of  an  arc  subtended  by 
a  given  chord,  chords  are  drawn  cutting  the  given  chord, 
the  product  of  each  whole  chord  from  P  and  its  segment 
adjacent  to  P  is  constant. 

Proof:  Take  two  such  chords,  PA  and  PC;  draw  diameter  PX;  etc. 
Rt.  &  PST  and  PCX  are  similar.     (Explain.) 

47.  If  from  any  point  within  a  triangle  ABC,  perpendiculars  to  the 
sides  are  drawn  —  OR  to  A  B,  OS  to  BC,  OT  to  A  C, 

At?  +  BS*  +  CT2  =  BR2  +  CS2  +  AT2.     [Draw  OA,  OB,  OC.] 


ORIGINAL  EXERCISES 


185 


APT  and  PBS 


48.  If  two  chords  intersect  within  a  circle  and  at 
right  angles,  the  sum  of  the  squares  of  their  four  seg- 
ments equals  the  square  of  the  diameter. 

To  Prove  :  AP*  +  BP2  +  CP2  +  DP2  =  AR2. 
Proof:  Draw  EC,  AD,  RD.     Chord  BR  is  _L  to  AB 

(240). 

.-.  CD  is  II  to  BR  (?)     .-.  arc  EC  =  arc  RD 
.-.  chord  EC  =  chord  RD  (?).     Also  A  ARD  is  rt.  A 
Now  AR2  =  AD*  +  DR2 
But  AD2  =  AP2  +  PD2  and  DR2  =  etc. 

49.  The  perpendicular  fromvany  point  flf  an  arc 
upon  its  chord  is  a  mean  proportional  between  the 
perpendiculars  from  the  same  point  to  the  tangents 
at  the  ends  of  the  chord. 

To  Prove  :  PR  :  PT  =  PT  :  PS. 
Proof:  Prove  &  ARP  and  BTP  similar;  also 
Thus,  get  two  ratios  each  =  to  PA  :  PB. 

50.  If  each  of  three  circles  intersects  the  other 
two,  the  three  common  chords  meet  in  a  point. 

To  Prove:  AB,  LM,  RS  meet  at  0. 

Proof:  Suppose  AB  and  LM  meet  at  O.  Draw 
R  O  and  produce  it  to  meet  the  ©  at  X  and  X'. 
Prove  OX=OX'  (by  320).  .-.  X,  X',  S  are 
coincident. 

61.   In   an  inscribed  quadrilateral  the  sum  of  the  products  of  the 
two  pairs  of  opposite  sides  is  equal  to  the  product  of 
the  diagonals. 

Proof  :  Draw  DX  making  Z  CDX  =  ZADB;  &ADB 
and  CDX  are  sim.   (?)  ;   also   &BCD  and  ADX  (?). 

Hence  AB  •  DC  =  DB  -  XC  (?), 

Also  AD  •  EC  =  DB  .  AX  (?). 

Adding,  etc. 

52.  If  AB  is  a  diameter,  EC  and  AD  tangents,  meeting  chords  AF 
and  BF  (produced)  at  C  and  D  respectively,  AB  is  a  mean  proportional 
between  the  tangents  EC  and  AD. 

53.  If  from  a  point  A  on  the  circumference  of  a  circle  two  chords  are 
drawn  and  a  line  parallel  to  the  tangent  at  A  meet  them,  the  chords  and 
their  segments  nearer  to  A  are  inversely  proportional. 

ROBBINS'S    NEW    PLANE    GEOM.  —  13 


186  BOOK   III.     PLANE   GEOMETRY 

CONSTRUCTION  PROBLEMS 

PROPOSITION  XLII.    PROBLEM 
342.    To  find  a  fourth  proportional  to 


three  given  lines.  I 


? v B 


oxK 


Given :   Three  lines  «,  b,  c.  R 


\ 


a 


\ 


c  \ nz 


Required  :  To  find  a  fourth  propor-     A-          s  w 

tional  to  a,  5,  c. 

Construction:  Take  two  indefinite  lines,  AB  and  AC, 
•meeting  at  A.  On  AB  take  AE  =  to  a,  RV=  to  b.  On  AC 
take  AS  =  to  c.  Draw  ES. 

From  v  draw  VW  II  to  BS,  meeting  AC  at  W. 

Statement:    SJF  is  the  fourth  proportional  required.   Q.E.F. 

Proof:   In  AAVW,  RS  is  II  to  VW  (Const.). 

.-.  a:  b  =  c:  SW  (294). 

Q.E.D. 

PROPOSITION  XLIII.     PROBLEM 

343.  To  find  a  third  proportional  to  two  given  lines. 
Given:   (?). 

Required:   (?). 

Construction  :  . 

Like  that  for  342.  ^( 

Statement:   (?).     Proof:  (?)      A"":::T-:-£  —  -  —    ........  :C 

PROPOSITION  XLIV.     PROBLEM 

344.  To  divide  a  given  line  into  segments  proportional  to 
any  number  of  given  lines.  A       t      H  G        B 

Given:   AB;  a,  b,  c,  d. 


Required  :   To  divide  AB  into  parts     a  -  b  -,(..  d 
which  shall  be  proportional  to  a,  6,      c  _ 
o,  d. 


* 


CONSTRUCTION  PROBLEMS  187 

Construction:  Draw  AX  oblique  to  AB  from  A.  On  AX 
take  AC  =  to  #,  CD  =  to  6,  DE  —  to  c,  J£.F  =  to  d.  Draw  FB  also 
through  £",  D,  and  (7,  lines  II  to  FB,  as  #£,  DH,  and  CI. 

Statement:   Ai,  IH,  HG,  GB  are  the  required  parts.     Q.E.F. 
Proof  :   Ai:  a  =  IH:  b  =  HG  :  c  =  GB  :  d  (295). 

Q.E.D. 
PROPOSITION  XLV.     PROBLEM 

345.   To  construct  a  triangle  similar  to  a  given  triangle  and 
having  a  given  side  homologous  to  a  side  of  the  given  triangle. 
Given  :  A  ABC  and  RS  c 

yv  i  '•• 

homologous  to  ^4#.  /\. 

Required  :   To        con- 
struct a  A  on  RS  similar 


Construction:   At  R  construct  ^SRX=toZAi  at  s  con- 
struct Z  RST=  to  Z  J5,  the  sides  of  these  angles  meeting  at  T. 

Statement:   (?).  Proof:   (?).  (303). 

PROPOSITION  XLVI.     PROBLEM 

346.   To  construct  a  polygon  similar  to  a  given  polygon  and 
having  a  given  side  homologous  to  a  side  of  the  given  polygon. 

Given:  Polygon  EB\ 
line  A'B'  homologous 
to  AB. 

Required:  To  con- 
struct a  polygon  upon 
A*!*',  similar  to  poly- 
gon EB.  A  B 

Construction:   From  A  draw  diagonals  AC  and  AD. 

On  A'Bf  construct  A  A'B'C'  similar  to  A  ABC  (by  345). 

On  ArB'  construct  A  A'C'D'  similar  to  A  ACD.     Etc. 

Statement:   (?).  Proof:  (?).  (319). 


188  BOOK  III.    PLANE   GEOMETRY 

PROPOSITION  XLVII.     PROBLEM 

347 .  To  find  the  mean  proportional  D , 

between  two  given  lines.  /t 

Given :   Lines  a  and  b. 

Required:   To  find  the  mean  pro-    A1:"" A" 
portional  between  them.  b • — 

Construction :  On  an  indefinite  line,  AX,  take  AB  =  to  a  and 
BC  =  to  b.  Using  o,  the  midpoint  of  AC,  as  center,  and  AO  as 
radius,  describe  the  semicircle,  ADC.  At  B  erect  BD  _L  to 
AC,  meeting  the  arc  at  D.  Draw  AD  and  CD. 

Statement :  BD,  or  m,  is  the  mean  proportional  required. 

Q.E.F. 
Proof:  a:m  =  m:b.  (?)     Q.E.D. 

348.  A  line  is  divided  into  extreme  and  mean  ratio  if  one 
segment  is  a  mean  proportional  between  the  whole  line  and 
the  other  segment ;  in  other  words,  if  a  line  is  to  one  of  its 
parts  as  that  part  is  to  the  other  part.     (See  292.) 

PROPOSITION  XLVIII.     PROBLEM  ...?... 

349.  To  divide  a  line  into  ex- 
treme and  mean  ratio.  / 

•  ...£ 

Given:   Line  AB  =  a.  \    ...--'" 

Required:  To  divide  AB  into  ...-••*'"    \\ 

extreme  and  mean  ratio;  that  is,     £- — F'  a_x~    B 

so  that  AB  :  AF  =  AF:  FB.  • 

a 

Construction :  At  B  erect  BR,  -L  to  AB  and  =  to  AB.  Using 
C,  the  midpoint  of  BR,  as  center,  and  CB  as  radius,  describe 
a  O.  Draw  AC  meeting  O  at  D  and  E.  On  AB  take  AF  =  AD ; 
let  AF=  x. 

Statement :  F  divides  AB  so  that  AB  :  AF—  AF:  FB.  Q.E.F. 

Proof:  AB  is  tangent  to  O  C  (202). 

.-.  AE-  AD=AB2  (324). 


CONSTRUCTION  PROBLEMS  189 

Now                          AD  =  x  (187)  and  DE=a  (190). 

.-.  AE=  a  +  x  (Ax.  4). 

Substituting,  (a +x)x  =  a2  (Ax.  6). 
Or                       ax  +  x2  =  a2 

.-.  a  :x  =  x:a-x  (281). 

That  is,  AB  :  AF=AF:FB.  Q.E.D. 

PROPOSITION  XLIX.     PROBLEM. 

350.    To   divide  a  ...8... 

line    externally    into 

extreme    and    mean  / 

ratio.  '\  f*- 

Given:  (?).  ,""*\ 


Required:  (?).         F""                           A  B     % 
Construction :  The  same  as  in  349,  except  that  AFr  is  taken 
on  BA  produced,  =  to  AE. 

Statement :  AB  :  AF'  =  AFr  :  BF'.  Q.  E.  F. 

Proof:  AB  is  tangent  to  O  C  (202). 

.  •.  AE  :  AB  =  AB  :  AD  (325). 

. '.  AE  +  AB  :  AE  =  AB  +  AD  :  AB  (284). 

Now                                AE+AB  =  BFf  (Ax.  4). 

Also  AB  +  AD  =  AE=AF'  (Const.). 

Substituting,  BF1  :  AF1  =  AFf  :  AB  (Ax.  6). 

That  is,  AB:AFf  =  AFr  :  BFf .  Q.E.D. 

351.    The  lengths  of  the  several  lines  of  349  and  350  may 
be  found  by  algebra,  if  the  length  of  AB  is  known. 
Thus  if  AB  =  a,  we  know  in  349,  a  :  x  =  x :  a  —  x. 
Hence  x2  =  a2  —  ax.     Solving  this  quadratic, 
x  =  AF=^  a(V5  —  1)  ;   also,  a  —  x  =  BF=  %  a(3  —  V5). 
Likewise  in  350,  if  AB  =  a,  AF'  =  y,  a\y  =  y : a -f- y. 
Solving  for  y,y  =  AF'  =  |  a( V5  +  1). 
Also  a  +  y  =  BF'  =  J  a(3  +  V5). 


190  BOOK   III.     PLANE   GEOMETKY 

ORIGINAL  CONSTRUCTIONS 

It  is  required : 

1.  To  construct  a  fourth  proportional  to  lines  that  are  exactly  3  in., 
5  in.,  and  6  in.  long.     How  long  should  this  constructed  line  be? 

2.  To  construct  a  mean  proportional  between  lines  that  are  exactly 
4  in.  and  9  in.     How  long  should  this  constructed  line  be? 

3.  To  construct  a  fourth  proportional  to  three  lines  5  in.,  8  in.,  and 
10  in.  will  this  be  the  same  length  as  a  fourth  proportional  to   5  in., 
10  in.,  and  8  in.?  to  8  in.,  10  in.,  and  5  in.  ?  to  10  in.,  5  in.,  and  8  in.  ? 

4.  To  construct  a  third  proportional  to  lines  3  in.  and  6  in.  long. 

5.  To  produce  a  given  line  AB  to  point  P,  such  that  AB  :  AP  =3:5. 
[Divide  AB  into  three  equal  parts,  etc.] 

6.  To  divide  a  line  8  in.  long  into  two  parts  in  the  ratio  of  5:7. 
[Divide  the  given  line  into  12  equal  parts.] 

7.  To  solve  Ex.  6  by  constructing  a  triangle.     [See  297.] 

8.  To  divide  one  side  of  a  triangle  into  segments  proportional  to  the 
other  two  sides. 

9.  To  divide  one  side  of  a  triangle  externally  into  segments  propor- 
tional to  the  other  sides. 

10.  To  construct  two  straight  lines  having  given  their  sum.  and  ratio. 
[Consult  Ex.  6.] 

11.  To  construct  two  straight  lines  having  given  their  difference  and 
ratio.     [Consult  Ex.  5.] 

12.  To  construct  a  triangle  similar  to  a  given  triangle  and  having  a 
given  perimeter.     [First,  use  344.] 

13.  To  construct  a  right  triangle  having  given  its  perimeter  and  an 
acute  angle.     [Constr.  a  rt.  A  having  the  given  acute  Z.     Etc.] 

14.  To  construct  a  triangle   having  given   its  perimeter  and  two 
angles.     [Constr.  a  A  having  the  two  given  A.     Etc.] 

15.  To  construct  a  triangle  similar  to  a  given  triangle  and  having  a 
given  altitude. 

16.  To  construct  a  rectangle  similar  to  a  given  rectangle  and  having 
a  given  base. 

17.  To  construct  a  rectangle  similar  to  a  given  rectangle  and  having 
a  given  perimeter. 

18.  To  construct  a  parallelogram  similar  to  a  given  parallelogram 
and  having  a  given  base. 


ORIGINAL   CONSTRUCTIONS 


191 


19.  To  construct   a   parallelogram  similar  to  a  given  parallelogram 
and  having  a  given  perimeter. 

20.  To   construct  a  parallelogram  similar  to  a  given  parallelogram, 
and  having  a  given  altitude. 

21.  To  draw  through   a  given  point  another  line,  which  is  termi- 
nated by  the  outer  two  of  three  lines  meet- 
ing in   a  point  and   bisected   by  the  inner 

one. 

Construction:  From   E   on  BD  draw  Us. 
Etc.     Through  P  draw  RT  II  to  GF. 

Statement:  RS  =  ST. 


T    C 


22.  To  inscribe  in  a  given  circle  a  triangle  similar  to  a  given  triangle. 
Construction:  Circumscribe  a  O  about  the  given  A;  draw  radii  to  the 

vertices ;  at  center  of  given  O  construct  3  A  =  to  the  other  central  angles. 

23.  To  circumscribe  about  a  given  circle  a  triangle  similar  to  a  given 
triangle. 

Construction :  First,  inscribe  a  A  similar  to  the  given  A. 

24.  To  construct  a  right  triangle,  having  given 
one  leg  and  its  projection  upon  the  hypotenuse. 

25.  To  inscribe  a  square  in  a  given  semicircle. 
Construction:  At  B  erect  BD  JL  to  AB  and 

=  to  AB-,  draw  DC,  meeting  O  at  R ;  draw  R U  II 
to  BD.    Etc. 

Statement:  (?). 

Proof:  RSTU  is  a  rectangle  (?).  A  CRU  is  similar  to  A  CDB  (?). 
/.  CU :  CB  =  UR:BD  (?).  But  CB  =  \BD  (?).  :.CU=%UR  (?).  Etc. 

26.  To  inscribe  in  a  given  semicircle  a  rectangle  similar  to  a  given 
rectangle. 

Construction :  From  the  midpoint  of  the  base  draw  line  to  one  of  the 
opposite  vertices.  At  given  center  construct  an  Z  =  to  the  Z  at  the  mid- 
point. Proceed  as  in  Ex.  25. 

27.  To  inscribe  a  square  in  a  given  triangle. 
Construction :  Draw  altitude  A  D ;  construct 

the  square  ADEF  upon  AD  as  a  side;  draw 
BF  meeting  A  C  at  72. 

Draw  RU  II  to  AD ;  RS  II  to  BC.     Etc. 


192  BOOK   III.     PLANE   GEOMETRY 

28.  To  inscribe  in  a  given  triangle  a  rectangle  similar  to  a  given 
rectangle. 

Construction  :  Draw  the  altitude.    On  this  construct  a  rectangle  similar 
to  the  given  rectangle. 
Proceed  as  in  Ex.  27. 

29.  To  construct  a  circle  which  shall  pass 
through  two  given  points  and  touch  a  given 
line. 

Given  :  Points  A  and  B  ;  line  CD.  f*  . 


P  R 

Construction:    Draw   line  AB  meeting  CD 

at  P.  Construct  a  mean  proportional  between  PA  and  PR  (by  347). 
On  PD  take  PR  =  to  this  mean.  Erect  OR  _L  to  CD  at  R,  meeting  _L 
bisector  of  AB  at  0.  Use  0  as  center,  etc. 

30.  To  construct  a  line  =  to  V2  in.     [Diag.  of   square   the  side  of 
which  is  1  in.] 

31.  To  construct  a  line  =  to  A/5  in. 

[Hyp.  of  a  rt.  A,  whose  legs  are  1  in.  and  2  in.  respectively.] 

32.  To  divide  a  line  into  segments  in  the  ratio  of  1  :  A/2. 

33.  To  divide  a  line  into  segments  in  the  ratio  of  1  :  A/5. 

34.  To  construct  a  line  x,  if  x  =  —  ,  and  a,  6,  c  are  lengths  of  three 

c 
given  lines.     [That  is,  to  construct  x,  if  c  :  a  =  b  :  x  (281)  .] 

35.  To  construct  a  line  x,  if  x  =  —  .     [3  c  :  a  =  b  :  a:.] 

3c 

36.  To  construct  a  line  x,  if  x  =  Vab.     [a  :  x  =  x  :  &.] 

37.  To  construct  a  line  x,  if  x  =  —  . 

c 

38.  To  construct  a  line  a;,  if  x  =  Va2  —  62.     [a  +  b  :  x  —  x  :  a  —  &.] 

39.  To  construct  a  line  x,  if  x  =  —  . 

c 

40.  To  construct  a  line  y,  if  ay  =  f  62.     . 

41.  To  construct  a  line  =  to  A/10  in. 

42.  To  construct  a  line  =  to  2  A/6  in. 

43.  To  construct  a  line  =  to  Va2  +  62,  if  a  and  &  are  given  lines. 


BOOK  IV 


AREAS   OF  POLYGONS 
THEOREMS   AND   DEMONSTRATIONS 

352.  The  unit  of  surface  is  a  square  each  side 
of  which  is  a  unit  of  length. 

353.  The  area  of  a  surface  is  the  number  of 
units   of  surface  it  contains.     The  area  of  a 
surface  is  the  ratio  of  that  surface  to  the  unit 
of  surface. 


UNIT   OF    LENGTH 


NOTE.  It  is  often  convenient  to  speak  of  "triangle,"  "rectangle," 
etc.,  when  one  really  means  "the  area  of  a  triangle,"  or  "the  area  of  a 
rectangle,"  etc. 

PROPOSITION  I.     THEOREM 

354.  If  two  rectangles  have  equal  altitudes,  they  are  to  each 
other  as  their  bases. 

Given:  Rectangles  AC  and 
EG  having  equal  altitudes, 
with  bases  AB  and  EF. 

To  Prove : 

AC:EG  =  ABiEF. 

Proof:  I. 


ABE  F 

If  AB  and  EF  are  commensurable. 
There  is  a  common  unit  of  measure  of  AB  and  EF  (225). 
Suppose  this  unit  is  contained  3  times  in  AB  and  5  times  in 
EF.     Hence  AB:EF=3:5  (Ax.  3). 

Draw  lines  through  these  points  of  division  J_  to  the  bases. 
These  divide  rectangle  AC  into  three'  parts  and  EG  into  5 
parts,  and  all  of  these  eight  parts  are  equal  (134). 

Hence  A  C :  EG  =  3  : 5  (Ax.  3). 

.'.  AC:EG  =  AB'.EF      (Ax.  1).  Q.E.D. 

193 


194 


BOOK   IV.     PLANE   GEOMETRY 


II.  If  AB  and  EF  are  incom- 
mensurable. 

There  does  not  exist  a  com- 
mon unit  (225). 

Divide  AB  into  several  equal 
parts.  Apply  one  of  these  as  a 
unit  of  measure  to  EF. 

There  is  a  remainder,  JtF 

Draw  RS  _L  to  EF. 


C     H 


SG 


AB 


(Hyp.)- 
(Case  I). 


Now  —  = 

ES       ER 

Indefinitely  increase  the  number  of  equal  parts  of  AB  ; 
that  is,  indefinitely  decrease  each  part,  or  the  unit  or  di- 
visor. Then  the  remainder,  RF,  is  indefinitely  decreased. 

That  is,  RF  approaches  zero  as  a  limit, 

Also  RFGS  approaches  zero  as  a  limit. 

Hence  ER  approaches  EF  as  a  limit, 

Also  ES  approaches  EG  as  a  limit. 

A  C1  AC* 

Therefore          —  approaches  —  as  a  limit, 
ES  EG 

A  7?  A  7? 

Also  -  approaches  —  as  a  limit. 

ER  EF 


(229).     Q.E.D. 


AC  _AB 

'  EG  ~~  EF 

355.  COROLLARY.     Two  rectangles  having  equal  bases  are  to 
each  other  as  their  altitudes.     (Explain.) 

PROPOSITION  II.     THEOREM 

356.  Any  two  rectangles  are  to  each  other  as  the  products 
of  their  bases  by  their  altitudes. 


1  

X 

B 

A 

2 

i 

d 

AREAS 


195 


Given:  Rectangles  A  and  B  the  altitudes  of  which  are 
a  and  c,  and  the  bases  b  and  c?,  respectively. 

To  Prove  :  A-.B  =  a-b:c-  d. 

Proof  :  Construct  a  third  rectangle  X,  whose  base  is  b  and 
whose  altitude  is  c. 


Then 

Also 

Multiplying, 


-  =  - 
B      d 

A=a^b 

B     c  •  a 


(355). 

(354). 

(Ax.  3).     Q.E.D. 


PROPOSITION  III.     THEOREM 

357.  The  area  of  a  rectangle 
is  equal  to  the  product  of  its 
base  by  its  altitude. 

Given:    Rectangle    #,    with  i!  U  ! 

base  b  and  altitude  h. 

u  • 

To  Prove :  Area  of  R  =  b  •  h. 

Proof:    Draw  a  square  tr,  each  side  of   which  is  a  unit 
of  length.     This  square  is  a  unit  of  surface  (352). 

Now  |=^4  =  6.  A  (356). 


But 


-  =  the  area  of  R 
U 


the  area  of  R  =  b  •  h 


358.   COROLLARY. 
square  of  its  side. 


(353). 

(Ax.  1). 
Q.E.D. 

The  area  of  a  square  is  equal  to  the 

(357.) 


.  Ex.  I  have  enough  material  to  build  1000  yards  of  fence.  If  I  put 
this  around  a  square  field,  how  many  square  yards  will  the  field  contain  ? 
If  I  put  it  around  a  rectangular  field  that  is  four  times  as  long  as  it  is  wide, 
how  many  square  yards  will  the  field  contain  ? 


196  BOOK  IV.     PLANE   GEOMETRY 

PROPOSITION  IV.     THEOREM 

359.   The  area  of  a  parallelogram  is  equal  to  the  product  of 
its  base  by  its  altitude.  F     D  EC 

Given:   HJ ABCD,  with  base  b  and 
altitude  h. 


To  Prove :  Area  of  ABCD  —  b  •  h.          A  B 

Proof:  From  A  and  B,  the  extremities  of  the  base,  draw 
Js  to  the  upper  base  meeting  it  in  F  and  E  respectively. 
In  rt.  A  ADF  and  BCE, 

AF=BE  (124). 

AD=BC  (124). 

.  •.  A  ADF  is  congruent  to  A  BCE  (84). 

Now   from    the    whole   figure    subtract   A  ADF  and   the 

parallelogram  ABCD  remains.     And  from  the  whole  figure 

subtract  A  BCE  and  rectangle  ABEF  remains. 

.*.  O  ABCD  =  rectangle  ABEF  (Ax.  2). 

But         rectangle  ABEF  =  b  •  h  (357). 

.  •.  H  ABCD  =  b  -  h  (Ax.  1). 

Q.E.D. 

360.  COROLLARY.    All  parallelograms  having  equal  bases  and 
equal  altitudes  are  equal  in  area. 

361.  COROLLARY.    Two  parallelograms  having  equal  altitudes 
are  to  each  other  as  their  bases. 

Proof :  P  =  b  -  h  and  P1  =  b'  .  h  (359). 

Dividing,  p=^j=4  (Ax'3> 

Q.E.D. 

362.  COROLLARY.     Two  parallelograms  having  equal  bases 
are  to  each  other  as  their  altitudes.     Proof :  (?). 

363.  COROLLARY.     Any  two  parallelograms  are  to  each  other 
as  the  products  of  their  bases  by  their  altitudes.     Proof :  (?). 


AREAS  197 

PROPOSITION  V.     THEOREM 

364.  The  area  of  a  triangle  is  equal 
to  half  the  product  of  its  base  by  its 
altitude. 

Given:   A^i?C,with  base  b  and  alti- 
tude h. 

To  Prove  :  Area  of  A  ABC  =  |  b  •  h. 

Proof  :  Through  A  draw  AR  II  to  BC  and  through  C  draw 
CR  ||  to  AB,  meeting  AR  at  R. 

Now  ABCR  is  a  O  (120). 

The  area  of          O  ^£CE  =  b  -  h  (359). 

Dividing  by  2,  |  O  ^tfOR  =  |  b  •  A  (Ax.  3). 

Also  ±EJABCR  =  AABC  (126). 

.*.  the  area  of  A  ^ijBC  =  ^  6  •  ^  (Ax.  1).     Q.E.D. 

365.  COROLLARY.    A  triangle  having  the  same  base  and  alti- 
tude as  a  parallelogram  equals  half  the  parallelogram. 

366.  COROLLARY.    All  triangles  having  equal  bases  and  equal 
altitudes  are  equal  in  area. 

367.  COROLLARY.    All  triangles  having  the  same  base  and 
whose  vertices  are  in  a  line  parallel  to  the  base  are  equal. 

368.  COROLLARY.    Two  triangles  having  equal  altitudes  are 
to  each  other  as  their  bases. 

Proof:  Ar=i&  .  h  ;     andAr'  =  J&'A  (364). 


369.  COROLLARY.    Two  triangles  having  equal  bases  are  to 
each  other  as  their  altitudes.    Proof  :  (?). 

370.  COROLLARY.    Any  two  triangles  are  to  each  other  as  the 
products  of  their  bases  by  their  altitudes.    Proof  :  (?). 

371.  COROLLARY.    The  area  of  a  right  triangle  is  equal  to 
half  the  product  of  the  legs.    Proof  :  (?). 


198  BOOK  IV.    PLANE   GEOMETEY 

PROPOSITION  VI.     THEOREM 

372.  The  area  of  a  trapezoid  is  equal  to  half  the  product 
of  the  altitude  by  the  sum  of  the  bases. 

Given:  Trapezoid  ABCD,  with   alti- 
tude h  and  bases  b  and  c. 

To  Prove :   Area  =  \  h  •  (6  +  0-  A 

Proof:  Draw    diagonal    AC.       The 
A  ABC  and  A  DC  have  the  same  altitude,  7i,  and  their  bases 
are  b  and  <?,  respectively. 

Now  A  ABC  =  -J  6  •  h  (364). 

Also  A  ADC—  \c  •  h  (?). 

Adding,  &ABC  +  &ADC=\b  -  h  +  ±c  .  h  (Ax.  2). 

That  is,    trapezoid  ABCD  =  J  ^ .  (b  +  c)       (Ax.  6).     Q.E.D. 

373.  COROLLARY.     The  area  of  a  trapezoid  is  equal  to  the 
product  of  the  altitude  by  the  median. 

Proof :  Area  ABCD  =  J-  h  •  (b  +  c)  =  h  •  j  (b  +  c)         (372). 
But  ±(b  +  c)=  median  (138). 

Hence       area  of  trapezoid  A  BCD  =  h ••  m    (Ax.  6).    Q.E.D. 


Ex.  1.   If  one  parallelogram  has  half  the  base  and  the  same  altitude 
as  another,  the  area  of  the  first  is  half  the  area  of  the  second. 

Ex.  2.   If  one  parallelogram  has  half  the  base  and  half  the  altitude  of 
another,  its  area  is  one  fourth  the  area  of  the  second. 

Ex.  3.    State  and  prove  two  analogous  theorems  about  triangles. 

Ex.  4.   If  a  triangle  has  half  the  base  and  half  the  altitude  of  a  paral- 
lelogram, the  triangle  is  one  eighth  of  the  parallelogram. 
•     Ex.  6.   The  area  of  a  rhombus  equals  half  the  product  of  its  diagonals. 

Ex.  6.   The  diagonals  of  a  parallelogram  divide  it  into  four  triangles 
of  equal  areas. 

Ex.  7.   The  diagonals  of  a  trapezoid  divide  it  into  four  triangles,  two 
of  which  are  similar  and  the  other  two  have  equal  areas. 

Ex.  8.   If  a  parallelogram  has  half  the  base  and  half  the  altitude  of  a 
triangle,  its  area  is  half  the  area  of  the  triangle. 

Ex.  9.   The  line  joining  the  midpoints  of  two  sides  of  a  triangle  forms 
a  triangle  whose  area  is  one  fourth  the  area  of  the  original  triangle. 


AREAS  199 

Ex.  10.  The  line  joining  the  midpoints  of  two  adjacent  sides  of  a 
parallelogram  cuts  off  a  triangle  whose  area  is  one  eighth  of  the  area  of 
the  parallelogram. 

Ex.  11.    If  one  diagonal  of  a  quadrilateral  bi-     A«-— B 

sects  the  other,  it  also  divides  the  quadrilateral 
into  two  triangles  having  equal  areas. 

To  Prove :  A  ABC  =  A  ADC.  D 

Ex.  12.  Either  diagonal  of  a  trapezoid  divides  the  figure  into  two  tri- 
angles the  ratio  of  which  is  equal  to  the  ratio  of  the  bases  of  the 
trapezoid.  Prove  in  two  ways. 

Ex.  13.  If,  in  triangle  ABC,  D  and  E  are  the  midpoints  of  sides  AB 
and  A  C  respectively,  &BCD  =  A  EEC. 

Ex.  14.  If  the  diagonals  of  quadrilatefal  ABCD  meet  at  E,  and 
A  ABE  is  equal  in  area  to  A  CDE,  the  sides  AD  and  EC  are  parallel. 


PROPOSITION  VII.     THEOREM 

374.  If  two  triangles  have  an  angle  of  one  equal  to  an  angle 
of  the  other,  they  are  to  each  other  as 
the  products  of  the  sides  including  the 
equal  angles.  ^^"'/   \H 

Given  :  A  ABC  and  DEF,  /-  A  =  Z  D.      B 

To  Prove:     *A™=*]^A£. 

A  DEF       DE  •  DF 

Proof  :  Superpose  A  ABC  upon  A  DEF  so  that  the  equal  A 
coincide  and  BC  takes  the  position  denoted  by  GH.  Draw  GF. 

Now  A  DGH  and  DGF  have  the  same  altitude  (a  _L  from  G 
to  DF),  and  A  DGF  and  DEF  have  the  same  altitude  (a  _L  from 

FtoDE).  A  DGH       DH  ,        ADGF       DG  SOZQ^ 

.*.  -  =  —      and       -  =  —  (obo). 

A  DGF      DF  A  DEF      DE 


iv/r    n.-    i    •                             -                DG  •  DH  ,  *        Q\ 

Multiplying,                     _=_  (Ax.  3). 
A  DEF       DE  -  DF 

rp,    ,  .                     AABC_  AB  -  AC  (Ax.  6). 
i  natj  is* 

A  DEF       DE  -  DF  Q.E.D. 


200 


BOOK   IV.     PLANE   GEOMETRY 


Ex.  1.  If  two  triangles  have  an  angle  of  one  the 
supplement  of  an  angle  of  the  other,  the  triangles 
are  to  each  other  as  the  products  of  the  sides  includ- 
ing these  angles. 

C  B A 

Ex.  2.   If  two  triangles  of  equal  area  have  an 

angle  of  one  equal  to  an  angle  of  the  other,  the  sides  including  these 
angles  are  reciprocally  proportional. 

Ex.  3.  Any  two  sides  of  a  triangle  are  reciprocally  proportional  to 
the  altitudes  upon  them. 

Ex.  4.  In  triangles  of  equal  area  the  bases  and  the  altitudes  upon  them 
are  reciprocally  proportional. 

Ex.  6.  If  two  isosceles  triangles  have  the 
legs  of  one  equal  to  the  legs  of  the  other,  and 
the  vertex  angle  of  the  one  the  supplement  of 
the  vertex  angle  of  the  other,  the  triangles  have 
equal  areas. 

Ex.  6.  Two  triangles  are  equal  in  area  if  they 
have  two  sides  of  one  equal  to  two  sides  of  the 
other  and  the  included  angles  supplementary. 

Ex.  7.  The  area  of  a  triangle  is  equal  to 
half  the  perimeter  of  the  triangle  multiplied 
by  the  radius  of  the  inscribed  circle. 

Ex.  8.  The  area  of  a  polygon  circumscribed 
about  a  circle  is  equal  to  half  the  product  of 
the  perimeter  of  the  polygon  by  the  radius  of  the  circle. 

Ex.  9.  The  line  joining  the  midpoints  of  the  bases  of  a  trapezoid 
bisects  the  area  of  the  trapezoid. 

Ex.  10.  Any  line  drawn  through  the  midpoint  of  a  diagonal  of  a 
parallelogram,  intersecting  two  sides,  bisects  the  area  of  the  parallelo- 
gram. 

Ex.  11.  The  lines  joining  (in  order)  the  midpoints  of  the  sides  of  any 
quadrilateral  form  a  parallelogram  whose  area  is  half  the  area  of  the 
quadrilateral. 

Ex.  12.  If  any  point  within  a  parallelogram  is  joined  to  the  four  ver- 
tices, the  sum  of  one  pair  of  opposite  triangles  is  equal  to  the  sum  of  the 
other  pair;  that  is,  to  half  the  parallelogram. 


AREAS 


201 


Ex.  13.  Is  a  triangle  bisected  by  an  altitude?  by  the  bisector  of  an  angle? 
by  a  median  ?  by  the  perpendicular  bisector  of  a  side  ?     Give  reasons. 

Ex.  14.  If  the  three  medians  of  a  triangle 
are  drawn,  there  are  six  pairs  of  triangles 
formed,  one  of  each  pair  being  double  the  other. 

For  instance,  A  A  OB  =  2  A  A OE  ;  etc. 

Ex.  15.    If  the  midpoints  of  two  sides  of  a  triangle  are  joined  to  any 
point  in  the  base,  the  quadrilateral  formed  is  half  the  original  triangle. 

Ex.  16.  If  lines  are  drawn  from  the  mid- 
point  of  one  leg  of  a  trapezoid  to  the  ends  of 
the  other  leg,  the  middle  triangle  thus  formed 
is  equivalent  to  half  the  trapezoid. 

Ex.  17.    The  area  of  a  trapezoid  is  equal  to    A  "~° 

the  product  of  one  of  the  non-parallel  sides,  by  the  perpendicular  upon  it 
from  the  midpoint  of  the  other. 

Ex.  18.  If  through  the  midpoint  of  one  of 
the  non-parallel  sides  of  a  trapezoid  a  line  is 
drawn  parallel  to  the  other  side,  the  parallelo- 
gram formed  is  equivalent  to  the  trapezoid. 

Ex.  19.    The  sum  of  the  three  perpendicu- 
lars  drawn  to  the  three  sides  of  an  equilateral  triangle  from  any  point 
within  is  constant  (being  equal  to  the  altitude  of  the  triangle). 

Proof:  Join  the  point  to  the  vertices.     Set  the  sum  of  the  areas  of  the 
three  inner  A  equal  to  the  area  of  the  whole  A.     Etc. 

Historical  Note.     Sir  Isaac  Newton  was  born  on  Dec.  25,  1642,  at 
Grantham,    England.     At  an  early  age   he  exhibited  a  fondness  and 
aptitude  for  mechanical  contrivances,  —  windmills, 
water-clocks,  kites  and  dials. 

Later  in  his  career  he  studied  Descartes'  geome- 
try and  was  inspired  with  a  love  for  all  the  math- 
ematical studies.     In  the  years  1665  and  1666  he 
made  many   important  mathematical   discoveries. 
He  invented  improvements  for  both  the  telescope 
and  microscope,  and  discovered  the  existence  of  the 
spectrum.     The  study  of  Kepler's  laws  of  motion 
resulted  in  Newon's  discovery  of  the  law  of  gravita- 
tion, which  he  discussed  with  such  contemporaries  as  Sir  Christopher 
Wren,  Hooke,  and  the  astronomer  Halley.     He  also  discovered  the  bi- 
nomial theorem,  which  was  inscribed  on  his  tomb  when  he  died  in  1727. 
He  has  always  been  honored  as  the  greatest  mathematician  of  all  time. 

ROBBINS'8    NEW   PLANE    GEOM. 14 


NEWTON 


202 


BOOK   IV.     PLANE   GEOMETRY 


PROPOSITION  VIII.     THEOREM 

375.   Two  similar  triangles  are  to  each  other  as  the  squares 
of  any  two  homologous  sides. 

A  .D 


Given:  Similar  A  ABC  and  DEF. 

To  Prove:      AABC^AB^  ==^  =  g^> 

A  DEF      DE2      ^>F2      EF* 

Proof :  Denote  a  pair  of  homologous  altitudes  by  h  and  h1 ', 
and  the  corresponding  bases  by  &  and  &'. 
AABC_  b  -  h  _b      h_ 

~v'n 


Now 
But 

Substituting, 
That  is, 


A  DEF     b'  -  h' 

*L  =  L 
h'~b' 

AABC  =  b     b_ 
ADEF~b'    b1 


b* 


or-So*. 


ADEF    EF*         DE* 


(311). 
(Ax.  6). 

Q.E.D. 


DF 


PROPOSITION  IX.     THEOREM 

376.  Two  similar  polygons  are  to  each  other  as  the  squares  of 
any  two  homologous  sides  and  as  the  squares  of  their  perimeters. 


Given:  Similar  polygons  ABCDE  and  A'B'C'D'E',  with  per- 
imeters  P  and  Pf  respectively. 


AREAS  203 

TV  D«-r  x  "Ay«u"  ABCDE  A!?  -,        P2 

To  Prove :  -— — ^  —  =  =    —  =  etc.,  and  = 

Polygon  A'B'C'D'E'     ~ArBr  P12 

Proof :  Draw  from  homologous   vertices,  A  and  A',  all  the 

pairs  of  homologous  diagonals,  dividing  the  polygons  into  A. 

These  A  are  similar  in  pairs.  (318). 

An  A  T?2 

•">  ^iJj  xr»rrr-\ 


A  S        CJ? 

^011^. 

(313). 
(287). 
(Ax.  1). 
(291). 
(Ax.  6). 

(Above). 
(Ax.  6). 

Q.E.D. 

(317). 
(287). 
(Ax.  1). 

Q.E.D. 

AS'     C'D'* 

A  T        Z>2? 

A  T'      D<E>* 
g   j.                           AB         BC         CD         DE 

A'B'     B'C'     C'D'     D'E' 

A   T>                      T>  n                     fiT\                      T\  I/  • 

-/1-O               t>(J              \jU              JJJL 

A  R       AS       AT 

'  A  R'      As'      AT' 
t    AR  +  AS   +  AT        AR 

'  A  R'  +  A  s'  +  A  T'      ARf 
Substituting        Polygon  ABCDE         A  R 

Polygon  A'B'C'D'E'     AR' 
But                              AR  __AB2 

Polygon  ABCDE           AB2 

Polygon  A'B'C'D'E'     J/^2 
Also                       P  :  P'  =  AB  :  A'  Bf  =  etc. 

.  -.  P2  :  P'2  =  AB*  :  A'  Br*  =  etc. 
Polygon  ABCDE          P2 

'  Polygon  A'B'C'D'E'     p'2 

204  BOOK   IV.     PLANE   GEOMETRY 

PROPOSITION  X.     THEOREM 

377.  The  square  described  upon  the  hypotenuse  of  a  right 
triangle  is  equal  in  area  to  the  sum  of  the  squares  described 
upon  the  legs. 


Given:  (?). 
To  Prove:  (?). 


L   D 

Proof:  Draw  CL  J_  to  AB,  meeting  AB  at  K  and  ED  at  L. 
Draw  BF  and  CE. 

Now  A  ACS,  ACG,  and  BCR  are  all  rt.A  (Hyp.) 

Hence          ACH  and  BCG  are  straight  lines.  (45.) 

Also  AELK  and  BDLK  are  rectangles.  (Def.) 

In  A  ABF  and  ACE,  AB  =  AE,  AF  =  AC  (122). 

Z  BAF=  Z  CAE 
(Each  is  composed  of  a  rt.  Z  plus  Z  CAB). 

.-.  AABF^AACE  (52). 

Also  A  ABF  and  square  AG  have  the  same  base,  AF,  and 
the  same  altitude,  AC. 

.  -.  square  AG  =  2  •  A  ABF  (365). 

Similarly,         rectangle  AKLE  =  2  •  A  ACE  (365). 

.'.  rectangle  AKLE  =  square  AG  (Ax.  1). 

By  drawing  AI  and   CD,  it  may  be  proved  in  the  same 

manner  that  rectangle  BDLK  =  square  BH. 

By  adding,  square  AD  =  square  AG  -f-  square  BH  (Ax.  2). 

Q.E.D. 


AEEAS 


205 


378.  COROLLARY.     The   square  described  upon  one  of  the 
legs  of  a  right  triangle  is  equal  in  area  to  the  square  described 
upon  the  hypotenuse  minus  the  square  described  upon  the 
other  leg. 

PROPOSITION  XI.    THEOREM 

379.  If  the  three  sides  of  aright  triangle 
are  the  homologous  sides  of  three  similar 
polygons,  the  polygon  described  upon  the 
hypotenuse  is  equal  in  area  to  the  sum  of 
the  two  polygons  described  upon  the  legs. 


Given:  (?). 
Proof : 

And 


To  Prove:  (?). 

S 


Z? 


AB 


Adding, 


&+  T=AC2+  BC2  ^_AB2=^ 
*  AB2         ~  AB*~~ 

Clearing  of  fractions,  B=  8  +  T  (Ax.  3).     Q.E.D. 

380.  COROLLARY.    If  the  three  sides  of  a  right  triangle  are 
the  homologous  sides  of  three  similar  polygons,  the  polygon 
described  upon  one  of  the  legs  is  equal  in  area  to  the  polygon 
described  upon  the  hypotenuse  minus  the  polygon  described 
upon  the  other  leg. 

381.  COROLLARY.     The  two   squares 
described  upon  the  legs  of  a  right  triangle 
are  to  each  other  as  the  projections  of 

the  legs  upon  the  hypotenuse.  A 

Proof :   Square  S  =  A  c2  Square  T  =  i?c2 

Square  S  =  AC2  =  AB  •  AP  _  AP 
Square  T      J^2      AB  -  BP      BP 


(Ax.  3;  333). 


206 


BOOK   IV.     PLANE   GEOMETRY 


382.  COROLLARY.  If  two  similar  poly- 
gons are  described  upon  the  legs  of  a  right 
triangle  as  homologous  sides,  they  are  to 
each  other  as  the  projections  of  the  legs 
upon  the  hypotenuse. 

Polygon  8  _  A(?  _AB 
Polygon  T      Be2      AB 


Proof: 


=  (376;  333). 

BP     BP      ^  J 


Ex.  1.  In  the  figure  of  377,  prove  that : 
(t)  Points  /,  C,  and  F  are  in  a  straight  line. 
(ii)   CE  and  BF  are  perpendicular. 
(iii)   A  G  and  BH  are  parallel. 
(iv)  &AEF=A  CGH  =--  A  BDI  =  &ABC. 

Ex.  2.  The  sum  of  the  squares  described  upon 
the  four  segments  of  two  perpendicular  chords  in 
a  circle  is  equal  to  the  square  described  upon  the 
diameter.  (Fig.  is  on  page  185.) 

Ex.  3.  The  square  described  upon  the  sum  of  two 
lines  is  equal  to  the  sum  of  the  squares  described  upon 
the  two  lines,  plus  twice  the  rectangle  of  these  lines. 

To  Prove :  Square  AE  =  m*  +  n2  +  2  mn. 

Ex.  4.    The  square  described  upon  the  difference  of    H 
two  lines  is  equal  to  the  sum  of  the  squares  described 
upon  the  two  lines  minus  twice  the  rectangle  of  these    G 
lines. 

To  Prove :  Square  AD  =  m*  +  n2  -  2  mn. 

Ex.  5.   A  and  B  are  the  extremities  of  a  diameter       ^ 
of  a  circle;  C  and  D  are  the  points  of  intersection  of 
any  third  tangent  to  this  circle  with  the  tangents  at 
A  arid  B  respectively.   Prove  that  the  area  of  ABDC 
is  equal  to  \AB  •  CD. 

Ex.  6.   If  the   four   points  midway   between   the 
center  and  vertices  of  a  parallelogram  are  joined  in 
order,  a  parallelogram  is  formed  similar  to  the  original  parallelogram ; 
its  perimeter  is  half  of  the  perimeter  of  the  original  figure ;  and  its  area 
is  one  quarter  of  the  area  of  the  original  figure. 


—  R- 
n      n 

C 

n 

m-n 

m-n         m-n 

m-n 

m-n 

n 

1 

5 

n                       n 
m 

FORMULAS 


207 


Ex.  7.  If  two  triangles  of  equal  area  have  the  same  base  and  lie  on 
opposite  sides  of  it,  the  line  joining  their  vertices  is  bisected  by  the  line 
of  the  base. 

Ex.  8.  What  part  of  a  right  triangle  is  the  quadrilateral  which  is  cut 
from  the  triangle  by  a  line  joining  the  midpoints  of  the  legs  ? 

Ex.  9.  From  M,  a  vertex  of  parallelogram  LMNO,  a  line  MPX  is 
drawn  meeting  NO  at  P  and  LO  produced,  at  X.  LP  and  NX  are  also 
drawn.  Prove  that  triangles  LOP  and  XNP  are  equal  in  area. 

FORMULAS 

PROPOSITION  XII.     PROBLEM 

383.  To  derive  the  formulas  for  the  altitude  and  the  area  of 
an  equilateral  triangle,  in  terms  of  its  side. 

Solution:  Let  each  side  =  a,  and  alti- 
tude =  h.     h2  =  a2_^  =  3as 

4     '    4 

<v 

.        h 
2 


(335). 


Area  =  ta  .  h  =  -a  .  -  \/3       (364). 
L  22 


H. 


PROPOSITION  XIII.     PROBLEM 

384.    To  derive  a  formula  for  the  area  of  a  triangle  in  terms 
of  its  sides. 

Given:  A  A BC,  having  sides  a,  £,  c. 

Required:  To  derive   a   formula   for 
its  area,  containing  only  a,  5,  and  c. 

Solution :  Draw  altitude  AD. 


A 
6>   |/> 


Now 
Also 


CD=bpa  = 


—i  & 
AD2  =  AC?  - 


//»2    [    ^2  /?^V 

•••h*  =  b*-(        2^— J 


B 
(339). 

(335). 
(Ax.  6). 


208  BOOK  IV.     PLANE  GEOMETRY 


Hence  W=  h+*±»=*  Hi- 


za       ) I  Xa 

Alsc     fl^aft+ga  +  ff-cg   2a6-a2-62  + 


(by  factoring). 


4  a2 

Then  it  is  evident  that  a+  b—  <?  =  2(s  —  c) 
and  a  —  b  +  c  =  2(s  —  6)  and   —  a  +  6  4-  c  =  2(«  —  a). 

Substituting  above,  A  =  J2«  •  2Q- a)  •  2(s  -  6)  .  2(s- c) 

4  a2 

2 


Now  area  A=-a»A=.- 

2  "2    a 


.'.  Area  of  A  =  Vs(s  —  a)  (s  —  6)  (s  —  c). 


EXERCISE.   Find  the  area  of  a  triangle  whose  sides  are  17,  25,  28. 

Here,  a  =  17,  b  =  25,  c  =  28,  5  =  35,  s  -  a  =  18,  s  -  b  =  10,  s  -  c  =  7. 

Area  =  V35  •  18  . 10  •  7  =  V72 .  52  •  22  •  32  =  210. 

Ex.  1.   Find  the  area  of  the  triangle  whose  sides  are  7,  10,  11. 

Ex.  2.   Find  the  area  of  the  triangle  whose  sides  are  8,  15,  17. 

Ex.  3.   Find  the  area  of  the  equilateral  triangle  whose  side  is  8. 

Ex.  4.   Find  the  side  of  the  equilateral  triangle  whose  area  is  121  v^. 

Ex.  6.   Find  the  area  of  the  equilateral  triangle  whose  altitude  is  10. 

PROPOSITION  XIV.     PROBLEM 

385.    To  derive  formulas  for  the  altitudes  of  a  triangle  in 
terms  of  the  three  sides. 


Solution :  Area  =  J  a^0  =  Vs(s  —  «)(«  —  6)(*  —  c). 

C-^,'1       i        ,          \/s(S  — «)(«—  6)(S  —  C).fc         V«(8— «)(«  — 6)(S  — < 

Similarly,  fe6=  — » — » ^;  ^c=  - 


FORMULAS 
PROPOSITION  XV.     PROBLEM 


209 


386.    To  derive  the  formula  for  the  radius  of  the  circle  in- 
scribed in  a  triangle,  in  terms  of  the  sides  of  the  triangle. 


B 


Solution : 

Adding, 
[Because 

Hence 


Area  of  A  AOB  =  J  c  -  r  ] 
area   of  A  AOC  =  J  b  -  r  \ 
area   of  A  BOG  =  |  a  •  r  } 
area   of  A  ABC  =  i  (#  -{-  5  -f-  c)r  =  SP. 
JO  + 6 +  <?)==  s.] 

area  of  A  ABC 


r  = 


s 


r  — 


—  a)(s  — 


—  c) 


PROPOSITION  XVI.     PROBLEM 
387.     To  derive  the  formula  for  the 
radius  of  the  circle  circumscribed  about 
a  triangle,  in  terms  of  the  sides  of  the 
triangle. 
Solution : 


•  ha  =  b  -  c 

R  =  — 
2h' 


(328), 


7   ._ 
fi  ^ 


—  c) 


(385). 


a  •  b  •  c 


Ex.  1.  Find  the  radius  of  the  circle  inscribed  in,  and  the  radius  of 
the  circle  circumscribed  about,  the  triangle  whose  sides  are  17,  25,  28. 

Ex.  2.  Find  for  triangle  whose  sides  are  11,  14,  17,  the  radii  of  the 
inscribed  and  circumscribed  circles. 


210  BOOK   IV.     PLANE   GEOMETBY 


ORIGINAL  EXERCISES    (NUMERICAL) 

1.  The  base  of  a  parallelogram  is  2  ft.  6  in.  and  its  altitude  is  1  ft. 
4  in.     Find  the  area.     Find  the  side  of  a  square  of  equal  area. 

2.  The  area  of  a  rectangle  is  540  sq.  m.  and  its  altitude  is  15  m. 
Find  its  base  and  its  diagonal. 

3.  The  base  of  a  rectangle  is  3  ft.  4  in.  and  its  diagonal  is  3  ft.  5  in. 
Find  its  area. 

4.  The  bases  of  a  trapezoid  are  2  ft.  1  in.,  and  3  ft.  4  in.,  and  the 
altitude  is  1  ft.  2  in.     Find  the  area. 

6.   The  area  of  a  trapezoid  is  736  sq.  in.  and  its  bases  are  3  ft.  and 
4  ft.  8  in.     Find  the  altitude. 

6.  The  area  of  a  certain  triangle  whose  base  is  40  rd.  is  3.2  A.   Find 
the  area  of  a  similar  triangle  whose  base  is  10  rd.     Find  the  altitudes  of 
these  triangles. 

7.  The  base  of  a  certain  triangle  is  20  cm.     Find  the  base  of  a  simi- 
lar triangle  four  times  as  large;  of  one  five  times  as  large;  twice  as 
large ;  half  as  large ;  one  ninth  as  large. 

8.  The  altitude  of  a  certain  triangle  is  12  and  its  area  is  100.     Find 
the  altitude  of  a  similar  triangle  three  times  as  large.     Find  the  base  of 
a  similar  triangle  seven  times  as  large.     Find  the  altitude  and  the  base  of 
a  similar  triangle  one  third  as  large. 

9.  The  area  of  a  polygon  is  216  sq.  m.  and  its  shortest  side  is  8  m. 
Find  the  area  of  a  similar  polygon  whose  shortest  side  is  10  m.     Find  the 
shortest  side  of  a  similar  polygon  four  times  as  large ;  one  tenth  as  large. 

10.  If  the  longest  side  of  a  polygon  whose  area  is  567  is  14,  what  is 
the  area  of  a  similar  polygon  whose  longest  side  is  12?    of   another 
whose  longest  side  is  21  ? 

11.  Find  the  area  of  an  equilateral  triangle  whose  sides  are  each  6  in. ; 
of  another  whose  sides  are  each  10v'3  ft. 

12.  Find  the  area  of  an  equilateral  triangle  whose  altitude  is  4  in. ; 
of  another  whose  altitude  is  18  dm. 

13.  The  area  of  an  equilateral  triangle  is  64  A/3.     Find  its  side  and  its 
altitude. 

14.  The  area  of  an  equilateral  triangle  is  90  sq.  m.     Find  its  altitude. 

16.    Find  the  side  of  an  equilateral  triangle  whose  area  is  equal  to  a 
square  whose  side  is  15  ft. 


ORIGINAL   EXERCISES  211 

16.  The  equal  sides  of  an  isosceles  triangle  are  each  17  in.  and  the 
base  is  16  in.     Find  the  area. 

17.  Find  the  area  of  an  isosceles  right  triangle  whose  hypotenuse  is 
2  ft.  6  in. 

18.  Find  the  area  of  a  square  whose  diagonal  is  20  m. 

19.  There  are  two  equilateral  triangles  whose  sides  are  33  and  56 
respectively.     Find  the  side  of  the  equilateral  triangle  equal  to  their  sum. 
Find  the  side  of  the  equilateral  triangle  equal  to  their  difference. 

20.  There  are  two  similar  polygons  two  of  whose  homologous  sides 
are  24  arid  70.     Find  the  side  of  a  third  similar  polygon  equal  to  their 
sum ;  the  side  of  a  similar  polygon  equal  to  their  difference. 

21.  What  is  the  area  of  the  right  triangle  whose  hypotenuse  is  29 
cm.  and  whose  short  leg  is  20  cm.  ? 

22.  The  base  of  a  triangle  is  three  times  the  base  of  a  triangle  equal 
in  area.     What  is  the  ratio  of  their  altitudes  ? 

23.  The  bases  of  a  trapezoid  are  56  ft.  and  44  ft.  and  the  non-paral- 
lel sides  are  each  10  ft.     Find  its  area.     Also  find  the  diagonal  of  a 
square  of  equal  area. 

24.  The  base  of  a  triangle  is  80  m.,  and  its  altitude  is  8  m.     Find  the 
area  of  the  triangle  cut  off  by  a  line  parallel  to  the  base  and  at  a  dis- 
tance of  3  m.  from  it.     Find  the  area  of  another  triangle,  cut  off  by  a 
line  parallel  to  the  base  and  6  m.  from  it. 

25.  The  bases  of  a  trapezoid  are  30  and  55, 
and  its  altitude  is  10.     If  the  non-parallel  sides 
are  produced  till  they  meet,  find  the  area  of  the 
less  triangle  formed. 

[The  A  are  similar.    .-.30:  55  =  a:  a: +  10.   Etc.] 

26.  The  diagonals  of  a  rhombus  are  2  ft.  and  70  in.     Find  the  area; 
the  perimeter ;  the  altitude. 

27.  The  altitude  (h)  of  a  triangle  is  increased  by  n  and  the  base  (Z>) 
is  diminished  by  x  so  that  the  area  remains  unchanged.     Find  x. 

28.  The  projections  of  the  legs  of  a  right  triangle  upon  the  hypote- 
nuse are  8  and  18.     Find  the  area  of  the  triangle. 

29.  In  triangle  A B C,  AB  is  5,  B C  is  8,  and  AB  is  produced  to  P,  mak- 
ing BP  =  6.    BC  is  produced  (through  B)  to  Q  and  PQ  is  drawn  so  that 
triangle  BPQ  is  equal  in  area  to  triangle  ABC.     Find  the  length  of  BQ. 


212  BOOK   IV.     PLANE   GEOMETRY 

30.  The  angle  C  of  triangle  ABC  is  right;  AC  =  5;  BC  =  12.     BA 
is  produced  through  A,  to  D  making  AD  =  4;  CA  is  produced  through 
^4,  to  E  so  that  triangle  A  ED  is  equal  in  area  to  triangle  ABC.     Find 
AE. 

31.  Find  the  area  of  a  square  inscribed  in  a  circle  whose  radius  is  6. 

32.  Find  the  side  of  an  equilateral  triangle  whose  area  is  25  V3. 

33.  Two  sides  of  a  triangle  are  12  and  18.     What  is  the  ratio  of  the 
two  triangles  formed  by  the  bisector  of  the  angle  between  these  sides  ? 

34.  The  perimeter  of  a  rectangle  is  28  m.  and  one  side  is  5  m.     Find 
the  area. 

35.  The  perimeter  of  a  polygon  is  5  ft.  and  the  radius  of  the  inscribed 
circle  is  5  in.     Find  the  area  of  the  polygon. 

In  the  following  triangles,  find  the  area,   the   three  altitudes,   the 
radius  of  the  inscribed  circle,  the  radius  of  circumscribed  circle : 

36.  a  =  13,  b  =  14,  c  =  15.  38.  20,  37,  51. 

37.  a  =  15,  b  =  41,  c  =  52.  39.   140,  143,  157. 

40.  The  sides  of  a  triangle  are  15,  41,  52.    Find  the  areas  of  the  two 
triangles  into  which  this  triangle  is  divided  by  the  bisector  of  the  largest 
angle. 

41.  Find  the  area  of  the  quadrilateral  ABCD  if  AE  =  78  m.,  BC  = 
104  m.,  CD  =  50  m.,  AD  =  120  m.,  and  AC  =  130  m. 

42.  One  diagonal  of  a  rhombus  is  TV  of  the  other  and  the  difference 
of  the  diagonals  is  14.     Find  the  area  and  the  perimeter  of  the  rhombus. 

43.  A  trapezoid  is  composed  of  a  rhombus  and  an  equilateral  triangle. 
Each  side  of  each  figure  is  16  inches.     Find  the  area  of  the  trapezoid. 

44.  Find  the  side  of  an  equilateral  triangle   equal  in  area  to   the 
square  whose  diagonal  is  15\/2. 

46.   Which  of  the  figures  in  Ex.  44  has  the  smaller  perimeter? 

46.  In  a  triangle  whose  base  is  20  and  whose  altitude  is  12,  a  line  is 
drawn  parallel  to  the  base,  bisecting  the  area  of  the  triangle.     Find  the 
distance  from  the  base  to  this  parallel. 

47.  Two  lines  are  drawn  parallel  to  the  base  of  a  triangle  whose  base 
is  30  and  altitude  18.     These  lines  divide  the  area  of  the  triangle  into 
three  equal  parts.     Find  their  distances  from  the  vertex. 

48.  Around  a  rectangular  lawn  30  yards  x  20  yards  is  a  drive  16  feet 
wide.     How  many  square  yards  are  there  in  the  drive  ? 


CONSTRUCTION  PROBLEMS 


213 


CONSTRUCTION  PROBLEMS 
PROPOSITION  XVII.     PROBLEM 
388.   To  construct  a  square  equal  to  the  sum  of  two  squares. 


Given :  (?).  Required :  (?).  Construction :  Construct  a  rt.  Z.E, 
with  sides  EX  and  EY.  On  EX  take  EF  =  to  AB ;  on  ET 
take  EG  =  to  CD.  Draw  FG.  On  FG  construct  square  T. 


Statement : 
Proof : 


T  =  B  +  s. 

T  =  R+  S 


PROPOSITION  XVIII.     PROBLEM 

389.   To  construct  a  square  equal  to 
the  sum  of  several  squares. 

Given:   Squares  whose  sides  are  a,  6, 

c,d. 

Required :  To  construct  a  square  =  to 


Construction :  Construct  a  rt.  Z.  whose  sides  are  equal  to  a 
and  b.  Draw  hypotenuse  EC.  At  B  erect  a  J_  =  to  c  and 
draw  hypotenuse  DC.  At  D  erect  a  _L  =  to  c?,  etc. 

Statement :  The  square  constructed  on  EC  =  the  sum  of 
the  several  given  squares.  Q.E.F. 

(334). 

(?>. 

=  a2  +  i2 +<?  +  #'  (?).     Q.E.D. 


214 


BOOK   IV.     PLANE   GEOMETKY 


PROPOSITION  XIX.     PROBLEM 

390.   To  construct  a  square  equal  to  the  difference  of  two 
given  squares. 


p 

fi 

S 

P 

^X. 

\                       BCD' 

T 

Given:   (?).    Required:   (?). 

Construction :  At  one  end  of  indefinite  line,  EX,  erect  EG  _L 
to  EX  and  =  to  CD  (a  side  of  the  smaller  square,  s).  Using  G 
as  center  and  AB  as  radius,  describe  arc  intersecting  EX  at  F. 

Draw  GF.     On  EF  construct  square  T. 

Statement:   T=R—S.     Q.E.F.  Proof:   (?). 

391.  To  construct  a  polygon  similar  to  two  given  similar  poly- 
gons and  equal  to  their  sum. 

Construction:  Like  388.     Proof:   (379). 

392.  To  construct  a  polygon  similar  to  two  given  similar 
polygons  and  equal  to  their  difference. 

Construction:   Like  390.     Proof:    (380). 


Ex.  1.   Construct  a  right  triangle  whose  area  shall  equal  the  area  of 
a  given  square. 

Ex.  2.   Construct  an  isosceles  triangle  whose  area  shall  equal  the  area 
of  a  given  square. 

Ex.  3.   Construct  an  isosceles  triangle  equal  in  area  to  a  given  right 
triangle. 

Ex.  4.   Construct  an  isosceles  triangle  equal  in  area  to  any  given 
triangle. 


CONSTRUCTION   PROBLEMS  215 


PROPOSITION  XX.     PROBLEM 

393.    To  construct  a  square  equal  in  area  to  a  given  paral- 
lelogram. 

D  EC     ,   - 


ZIF: 


M 


A  B     A1  B1  E'  X  B' 

Given:  (?).  Required:  (?)  Construction:  Construct  a 
mean  proportional  between  the  base,  AB,  and  the  altitude,  BE-, 
on  this  mean  proportional,  B'G,  construct  a  square,  M. 

Statement:   Square  M  =  parallelogram  P.  Q.E.F. 

Proof:  AB:  BrG=BrG:  BE  (Const.). 

...  WG?=  AB  -  BE  (280). 

But  B7^  =  square  M  (358). 

And  AB  .  BE  =  a  P  (859) . 

.  *.  square  M  =  O  p  (Ax.  1). 

Q.E.D. 

394.   To  construct  a  square  equal  in  area  to  a  given  triangle : 

Construct  a  mean  proportional  between  half  the  base  and 
the  altitude,  and  proceed  as  in  393. 


Ex.  1.   Construct  a  square  equal  in  area  to  a  given  right  triangle. 

Ex.  2.  Construct  a  square  equal  in  area  to  the  sum  of  any  two  tri- 
angles ;  equal  to  their  difference. 

Ex.  3.  Construct  a  square  equal  in  area  to  the  sum  of  two  parallelo- 
grams ;  equal  to  their  difference. 

Ex.  4.  Construct  a  right  triangle  equal  in  area  to  a  given  parallel- 
ogram, and  with  the  same  base. 

Ex.  5.  Construct  an  isosceles  triangle  equal  in  area  to  a  given  parallel- 
ogram, and  with  the  same  base. 

Ex.  6.  Construct  on  the  same  base  as  a  square,  an  isosceles  triangle 
equal  in  area  to  the  square. 

Ex.  7.  Construct  a  right  triangle  equal  in  area  to  the  sum  of  two 
given  squares. 


216  BOOK   IV.     PLANE   GEOMETRY 

PROPOSITION  XXI.     PROBLEM 
396.   To  construct  a  triangle  equal  to  a  given  polygon. 

E  E 


Given :  Polygon  AD.     Required :  To  construct  a  A  =  to  AD. 

Construction:   Draw  a  diagonal,  BD,  connecting  any  vertex 

)  to  the  next  but  one  (D).  From  the  vertex  between 
these  (C),  draw  CG  II  to  BD,  meeting  AB  prolonged,  at  G. 
Draw  DG.  Repeat  (2d  figure)  by  drawing  EG  and  DH  II  to 
EG,  then  EH.  Repeat  again  by  drawing  AE,  FI II  to  AE,  then  El. 

Statement:  A  IHE=  polygon  ABCDEF.  Q.E.F. 

Proof :    In  first  figure,  A  BGD  =  A  BCD  (367). 

Add  polygon  ABDEF=  polygon  ABDEF 

.-.  pentagon  AGDEF  =  polygon  ABCDEF  (Ax.  2). 

Also,  in  second  figure,  A  GEE  =  A  GDE  (367). 

Add  polygon  AGEF=.  polygon  AGEF 

.*.  quadrilateral  AHEF=  pentagon  AGDEF  (Ax.  2). 

Again,  A  AIE  =  A  AFE  (367). 

Add  A  AHE  =  A  ARE 

.'.    A  IHE  =  quad.  AHEF  (Ax.  2). 

Hence  A  IHE=  polygon  ABCDEF  (Ax.  1). 

Q.E.D. 

396.   To  construct  a  square  equal  to  a  given  polygon. 

First,  construct  a  A  =  to  the  polygon  (by  395). 

Second,  construct  a  square  =  to  the  A  (by  394). 


Ex.  1.   Tell  how  to  construct,  by  two  methods,  a  square  equal  to  a 
parallelogram. 

Ex.  2.    How  can  rectilinear  figures  be  added  into  a  single  figure  ? 


CONSTRUCTION  PROBLEMS  217 

PROPOSITION  XXII.     PROBLEM 

397.    To  construct  a  parallelogram  (or  a  rectangle)  equal  to 
a  given  square,  and  having : 

I.  The  sum  of  its  base  and  altitude  equal  to  a  given  line. 
II.  The  difference  of  its  base  and  altitude  equal  to  a  given  line. 


fLJ^: 

\L_ 

\, 

Gf 

A  B  C     G  D 

I.  Given :   Square  8  and  line  CD. 

Required :  To  construct  a  O  =  to  8 ;  base  +  altitude  =  CD. 

Construction:  On  CD  as  a  diameter  describe  a  semicircle. 
At  C  erect  CE  J_  to  CD  and  =  to  AB.  Through  E  draw  EF  II 
to  CD,  meeting  the  circumference  at  F.  Draw  FG  J_  to  CD. 
Take  G'D'  =  to  GD  and  draw  XY  II  to  G'D'  at  the  distance 
from  it  =  to  CG.  On  XY  take  HI  =  GD.  Draw  HGf  and  ID1 '. 

Statement:   O  G'D'IH=  &,  and  base  +  alt.  =  CD.        Q.E.F. 

Proof :  G'D'IH  is  a  O  (129). 

Also  GD.GC  =  FG2  (332). 

But  GD  -  GC  =  n  G'D'IH  (359). 

And  FG2  =  EC2  =  AB2  =  square  8     (124,  358). 

Substituting,        O  G'D' IH=  square  8  (Ax.  6). 

Also  G'D'  +  G'C'  =  CD  (Ax.  4).     Q.E.D. 


Historical  Note.  The  following  extract  from  Nicolay  and  Hay's  Life 
of  Abraham  Lincoln  should  be  of  interest  to  the  student: 

"  His  wider  knowledge  of  men  and  things  had  shown  him  a  certain 
lack  in  himself  of  the  power  of  close  and  sustained  reasoning.  To 
remedy  this  defect,  he  applied  himself,  after  his  return  from  Congress,  to 
such  works  upon  logic  and  mathematics  as  he  fancied  would  be  service- 
able. Devoting  himself  with  dogged  energy  to  the  task  in  hand,  he  soon 
learned  by  heart  six  books  of  the  propositions  of  Euclid,  and  he  retained 
through  life  an  intimate  knowledge  of  the  principles  they  contain." 

ROBBINS'S    NEW    PLANE    GEOM. 15 


218 


BOOK   IV.     PLANE   GEOMETRY 


II.  Given :   Square  S  and  line  CD. 

Required :   To  construct  a  O  =  to  S ;  base  —  altitude  =  to  CD. 


f\F,-— -. 

1     A 

I''   \o 


X... 


Construction  :  On  CD  as  diameter,  describe  a  O,  o.  Ate 
erect  CE  _L  to  CD  and  =  to  AB.  Draw  EFOG  meeting  O  at  F 
and  G.  Take  EfGr  =  to  EG  and  draw  XY  II  to  E'G'  at  a  distance 


from  it  =  to  EF.     On  XY  take  fll=  to  EG.     Draw 
Statement  :   O  E'G'IH  =  s,  and  base  -  alt.  =  CD. 


Proof 


But 
And 

Also 


E'G'IH  is  a  O 
EC  is  tangent  to  O  O 
.'.  EG  •  EF=  EC2 

EG  •  EF=  (U  E'G'IH 

EC2  =  AB2  =  square  S 
.'.  O  E'G'IH  =  square  8 
E'G'  -  E'F'  =FG  =  CD         (190). 


and  IG1. 

Q.E.F. 

(129). 
(202). 
(324). 
(359). 
(358). 
(Ax.  6). 

Q.E.D. 


398.    To  find  two  lines  whose  product  is  given : 
I.  If  their  sum  is  also  given. 
II.  If  their  difference  is  also  given 


I.  If  their  sum  is  also  given.  1    ,™  on_  n 

r  [The  same  as  397.1 

.  J    ' 


PROPOSITION  XXIII.     PROBLEM 

399.   To  construct  a  square  having  a  given  ratio  to  a  given 
square. 


.  D 


A"- 


Given :   Square  J2,  and  lines  g  and 


CONSTRUCTION  PROBLEMS  219 

Required:  To  construct  a  square  such  that  it  (the  un- 
known square)  :  square  B  =  h  :  g. 

Construction  :  On  an  indefinite  line  AY  take  AB  =  to  g,  and 
BC  =  to  h.  On  AC  as  diameter  describe  a  semicircle.  At  B 
erect  BD  _L  to  AC,  meeting  arc  at  D.  Draw  AD  and  CD.  On 
AD  take  DE=  to  a,  and  draw  EF  II  to  AC,  meeting  DC  at  F. 
Using  DF  =  x,  as  a  side,  construct  square  S. 

Statement  :  s  :  R  =  h  :  g.  Q.E.F. 

Proof  :  Z  ADC  is  a  rt.  £  (240). 

.-.!     =  *  (381). 


But  *  =  ^>  (294). 

a      ^1D 


==2 


(287). 


<*      9 

But  2^  =  square  S,  and  a2  =  square  R  (358). 

Substituting,     square  S  :  square  R  =  h  :  g  (Ax.  6). 

Q.E.D. 

PROPOSITION  XXIV.     PROBLEM. 

400.   To  construct  a  polygon  similar  to  a  given  polygon  and 
having  a  given  ratio  to  it. 

..  D 


Given:  (?).    Required:  (?).     Construction  and  Statement 
are  the  same  as  for  Proposition  XXIII. 

Proof:  Very  much  like  the  proof  of  Proposition  XXIII. 


220  BOOK   IV.     PLANE   GEOMETRY 

PROPOSITION  XXV.     PROBLEM 

401.   To  construct  a  polygon  similar  to  one  given  polygon 
and  equal  in  area  to  another. 


Given:  Polygons  R  and  s.     Required:  (?). 

Construction :  Construct  squares  Rr  =  R,  and  s1  =  8  (by 
396).  Find  a  fourth  proportional  to  a,  b,  and  AB.  This 
is  CD.  Upon  CD,  homologous  to  AB,  construct  T  similar  to  R. 

Statement:  T  =  8.  Q.E.F. 


Proof:                        f  =  ^  (3T6> 

-  =  —  (Const.). 
b       CD 

•••t  =  —  (287). 
*2     c& 

-••!=£  (Ax.l). 


Now  a2  =  B'  =  R  ;  a2  =  s'  =  S       (358  &  Const.). 

Substituting,  -=-  (Ax.  6). 

.-.  T=S  (Ax.  3). 

Q.E.D. 


NOTE.  By  means  of  this  proposition  we  are  able  to  maintain  the  size 
of  a  rectilinear  figure,  but  change  its  shape  to  any  desired  form.  Thus 
we  can  construct  an  equilateral  triangle  equal  to  a  given  square.  The 
pupil  will  explain. 


ORIGINAL   CONSTRUCTIONS  221 

ORIGINAL   CONSTRUCTIONS 

It  is  required : 

1.  To  construct  a  right  triangle  equal  to  a  given  parallelogram. 

2.  To  construct  a  square  equal  to  the  sum  of  two  given  parallelograms. 

3.  To  construct  a  square  equal  to  the  difference  of  two  given  paral- 
lelograms. 

4.  To  construct  a  square  equal  to  the  sum  of  several  given  right 
triangles. 

6.   To  construct  a  square  equal  to  the  sum  of  several  given  paral- 
lelograms. 

6.  To  construct  a  square  equal  to  the  sum  of  several  given  triangles. 

7.  To  construct  a  square  equal  to  the  sum  of  several  given  polygons. 

8.  To  construct   a   square    equal  to  the   difference   of    two   given 
polygons. 

9.  To  construct  a  square  equal  to  three  times  a  given  square ;  seven 
times  a  given  square. 

10.  To  construct  a  right  triangle  equal  to  the  sum  of  several  given 

triangles. 

11.  To  construct  a  right  triangle  equal  to  the  difference  of  any  two 
given  triangles  ;  of  any  two  given  parallelograms. 

12.  To  construct  a  square  equal  to  a  given  trapezoid ;  equal  to  a  given 
trapezium. 

13.  To  construct  a  square  equal  to  a  given  hexagon. 

14.  To  construct  a  rectangle  equal  to  a  given  triangle,  having  given 
its  perimeter. 

15.  To  construct  an  isosceles  right  triangle  equal  to  a  given  triangle. 

16.  To  construct  a  square  equal  to  a  given  rhombus. 

17.  To  construct  a  rectangle  equal  to  a  given  trapezium,  and  having 
its  perimeter  given. 

18.  To  find  a  line  whose  length  shall  be  \/2  units.     [See  388.] 

19.  To  find  a  line  whose  length  shall  be  V3  units. 

20.  To  find  a  line  whose  length  shall  be  VTi  units. 

21.  To  find  a  line  whose  length  shall  be  V7  units. 

22.  To  find  a  line  whose  length  shall  be  V30  units. 


222  BOOK   IV.     PLANE   GEOMETRY 

23.  To  construct  a  square  which  shall  be  f  of  a  given  square. 

24.  To  construct  a  square  which  shall  be  f  of  a  given  square. 

25.  To  construct  a  polygon  which  shall  be  f  of  a  given  polygon,  and 
similar  to  it. 

26.  To  construct  a  square  which  shall  have  to  a  given  square  the 
ratio  V3  : 4 ;  the  ratio  4  :  \/3. 

27.  To  draw  through  a  given  point,  within  a  parallelogram,  a  line 
which  shall  bisect  the  parallelogram. 

28.  To  construct  a  rectangle  equal  to  a  given  trapezoid,  having  given 
the  difference  of  its  base  and  altitude. 

29.  To  construct  a  triangle  similar  to  two  given  similar  triangles  and 
equal  to  their  sum. 

30.  To  construct  a  triangle  similar  to  a  given  triangle  and  equal  to  a 
given  square.     [See  401.] 

31.  To  construct  a  triangle  similar  to  a  given  triangle  and  equal  to  a 
given  parallelogram. 

32.  To  construct  a  square  having  twice  the  area  of  a  given  square. 
[Two  methods.] 

33.  To  construct  a  square  having  3|  times  the  area  of  a  given  square. 

34.  To  construct  an  isosceles  triangle  equal  to  a  given  triangle  and 
upon  the  same  base. 

35.  To  construct  a  triangle  equal  to  a  given  triangle,  having  the 
same  base,  and  also  having  a  given  angle  adjoining  this  base. 

36.  To  construct  a  parallelogram   equal  to   a  given  parallelogram 
having  the  same  base  and  also  having  a  given  angle  adjoining  the  base. 

37.  To  draw  a  line  that  shall  be  perpendicular  to  the  bases  of  a 
parallelogram  and  that  shall  bisect  the  parallelogram. 

38.  To  construct  an  equilateral  triangle  equal  to  a  given  triangle. 
[See  401.] 

39.  To  trisect  (divide  into  three  equal  parts)  the  area  of  a  triangle, 
by  lines  drawn  from  one  vertex. 

40.  To  construct  a  square  equal  to  f  of  a  given  pentagon. 

41.  To  construct  an  isosceles  trapezoid  equal  to  a  given  trapezoid. 

42.  To  construct  an  equilateral  triangle  equal  to  the  sum  of  two  given 
equilateral  triangles. 


ORIGINAL   CONSTRUCTIONS  223 

43.  To  construct  an  equilateral  triangle  equal  to  the  difference  of  two 
given  equilateral  triangles. 

44.  To  construct  upon  a  given  base  a  rectangle  that  shall  be  equal  to 
a  given  rectangle. 

Analysis :   Let  us  call  the  unknown  altitude  x. 
Then  b  •  h  =  V  •  x  (?).     Hence,  b':b  =  h:x  (?). 


A 

b 


That  is,  the  unknown  altitude  is  a  fourth  pro- 
portional to  the  given  base,  the  base  of  the  given  rec-  b' 
tangle,  and  the  altitude  of  the  given  rectangle. 

Construction :  Find  a  fourth  proportional,  x,  to  b',  b  and  h.  Construct 
a  rectangle  having  base  =  b'  and  alt.  =  x.  

Statement:  This  rectangle,  B  =  A.  x  B 

Proof:   bf  :  b  =  h  :  x   (Const.).     .-.  b'x=  bh  (?).     p 

But  Vx  =  the  area  of  B  (?),  etc. 

45.  To  construct  a  rectangle  that  shall  have  a  given  altitude  and  be 
equal  to  a  given  rectangle. 

46.  To  construct  a  triangle  upon  a  given  base  that  shall  be  equal  to 
a  given  triangle. 

47.  To  construct  a  triangle  that  shall  have  a  given  altitude  and  be 
equal  to  a  given  triangle. 

48.  To  construct  a  rectangle  that  shall  have  a  given  base,  and  shall 
be  equal  to  a  given  triangle. 

49.  To  construct  a  triangle  that  shall  have  a  given   base,  and  be 
equal  to  a  given  rectangle. 

50.  To  construct  a  triangle  that  shall  have  a  given  base,  and  be 
equal  to  a  given  polygon. 

51.  To  construct  the  problems  45,  46,  47,  48,  49,  substituting  "parallel- 
ogram "  in  each  case  for  the  figure  to  be  constructed. 

52.  To  construct  upon  a  given  hypotenuse,  a  right  triangle  equal  to  a 
given  triangle. 

53.  To  construct  upon  a  given  hypotenuse,  a  right  triangle  equal  to  a 
given  square. 

54.  To  construct  (a)  a  triangle  which  shall  have  a  given  base,  a  given 
adjoining  angle,  and  be  equal  to  a  given  triangle ;  (&)  a  triangle  equal  to 
a  given  'square ;  (c)  a  triangle  equal  to  a  given  polygon. 

56.  To  construct  (a)  a  parallelogram  which  shall  have  a  given  base,  a 
given  adjoining  angle,  and  be  equal  to  a  given  parallelogram ;  (5)  a  par- 
allelogram equal  to  a  given  triangle ;  (c)  a  parallelogram  equal  to  a  given 
polygon. 


224  BOOK   IV.     PLANE  GEOMETRY 

56.    To  construct  a  line,  DE,  from  D,  a  given  point  in  AB  of  triangle 
ABC,  so  that  DE  bisects  the  triangle. 

Analysis:  After  DE  is  drawn,  A  ABC  =  2  A  ADE 
(Hyp.).     But  A  ABC  :  A  ADE  =  AB  •  AC  : 
AD  •  AE  (?).     Hence,  AB  •  AC  =  2  (A D  -  AE)  (Ax.  6). 

.-.  2AD-.AB  =  AC:  x  (?).  Thus  a:  (that  is,  AE) 
is  a  fourth  proportional  to  three  given  lines. 

67.    To  draw  a  line  meeting  two  sides  of  a  triangle  and  forming  an 
isosceles  triangle  equal  to  the  given  triangle. 

Analysis:  Suppose  A X  is  a  leg  of  the  required  isos- 
celes A.  .-.  A  ABC  :&AXX  =  AB  -  AC  :  AX  •  AX'. 
But  the  A  are  equal  and  A X  =  AX'  (Hyp.).  y 

Hence,  AB  •  AC  —  AX2.    .-.  AX  is  a  mean  proper-^- \c 

tional  between  AB  and  A  C.  x- 

58.  To  draw  a  line  parallel  to  the  base   of  a  triangle  which  shall 
bisect  the  triangle, 

59.  To  draw  a  line  meeting  two  sides   of  a  triangle  forming   an 
isosceles  triangle  equal  to  half  the  given  triangle. 

60.  To  draw  a  line  parallel  to  the  base  of  a  triangle  forming  a  tri- 
angle equal  to  one  third  of  the  original  triangle. 

61.  To  draw  a  line  parallel  to   the  base  of  a 

trapezoid  so  that  the  area  is  bisected.  /'   \ 

Analysis :  A  OXX'  =  J  (A  OAD  +  A  OBC)  and 
is  similar  to  A  OB  C. 


62.   To  construct  two  lines  parallel  to  the  base 


of  a  triangle,  that  shall  trisect  the  area  of  the  triangle. 

63.  To  construct  a  triangle,  having  given  its  angles  and  its  area. 

Analysis :  The  required  A  is  similar  to  any  A  containing  the  given  A. 
The  given  area  may  be  a  square.     This  reduces  the  problem  to  401. 

64.  To  find  two  straight  lines  in  the  ratio  of  two  given  polygons. 


BOOK  V 

REGULAR  POLYGONS.     CIRCLES 
THEOREMS   AND   DEMONSTRATIONS 

402.  A  regular  polygon  is  a  polygon  which  is  equilateral 
and  equiangular. 

PROPOSITION  I.     THEOREM 

403.  An  equilateral  polygon  inscribed 
in  a  circle  is  regular. 

Given:  AG,  an  equilateral  inscribed 
polygon. 

To  Prove:  AG  is  regular. 

K^=:_ 

j 

Proof :  chord  AB  =  chord  EC  =  chord  CD  =  etc.  (Hyp.). 

.'.  arc  AB  =  arc  BC=  arc  CD  =  etc.  (196). 

.'.  arc  AC  =  arc  BD  =.arc  CE  =  etc.  (Ax.  3). 

.•.Z^LBC=Z£CD  =  Z  CDE=etc.  (239). 

That  is,  the  polygon  is  equiangular. 

.'.  the  polygon  is  regular       (402).    Q.  E.  D. 

404.  COROLLARY.     If  the  circumference  of  a  circle  is  divided 
into  any  number  of  equal  arcs,  and  the  chords  of  these  arcs 
are  drawn,  they  form  an  inscribed  polygon. 

Given :  Arc  AB  =  arc  BC  =  arc  CD  =  etc.  and  chords  AB,  etc. 
To  Prove :  Polygon  AG  is  a,  regular  polygon. 
Proof:  Chords  AB,  BC,  CD,  etc.  are  all  equal.  (197). 

.'.  the  polygon  is  regular  (403).   Q.  E.  D. 

405.  COROLLARY.     If  chords  are  drawn  joining  the  alternate 
vertices  of  an  inscribed  regular  polygon  (having  an  even  num- 
ber of  sides),  another  inscribed  regular  polygon  is  formed. 

225 


226 


BOOK   V.     PLANE   GEOMETRY 


PROPOSITION  II.     THEOREM 

406.  If  the  circumference  of  a  circle  is  divided  into  any 
number  of  equal  parts,  and  tangents  are  drawn,  at  the  several 
points  of  division,  they  form  a  circum- 
scribed regular  polygon. 

Given:  Arcs  AB,  BC,  CD,  etc.  all 
equal  ;  and  GH,  HI,  H,  etc.  tangents 
at  A,  B,  C,  etc. 

To  Prove  :  Polygon  HK  is  regular. 

Proof  :  Draw  chords  AB,  BC,  CD,  etc. 

In  A  ABH,  BCI,  CDJ,  etc.  AB  =  BC 
=  CD,  etc.  .  (197). 

Z  HAB  =  Z  HBA  =  Z  IBC  =  Z  ICB  =Z  JCD,  etc. 
.*.  these  A  are  congruent  and  isosceles 


That  is, 
Also 

That  is, 


(237). 
(76;  114). 
.-.  Ztf=Zi=Z</=etc.  (27). 

polygon  HK  is  equiangular. 
AH=HB  =  BI=IC=  CJ,  etc.       (24  or  206  ;  27). 
.-.  Hl=H=JK=etc.  (Ax.  3). 


polygon  HK  is  equilateral. 
.*.  polygon  HK  is  regular 


(402). 
Q.E.D. 


407.    COROLLARY.     If  the  circumfer- 
ence of   a  circle  is  divided  into  any    H 
number  of  equal  parts  and  tangents 
are    drawn  at  their   midpoints,  they   G 
form  a  circumscribed  regular  polygon. 

Given:  (?)     To  Prove:  (?).  R 


Proof:  Arcs  AB,  BC,  CD,  etc.  are  all  equal  (Hyp.). 

Also  arcs  AI,  IB,  BK,  KG,  CM,  etc.  are  all  equal     (Ax.  3). 

.-.  arcs  IK,  KM,  MO,  etc.  are  all  equal        (Ax.  3). 

Therefore  the  polygon  is  regular       (406).     Q.E.D. 


REGULAR  POLYGONS  227 

408.  COROLLARY.     If  the  vertices  of 
an  inscribed  regular  polygon  are  joined 
to  the  midpoints  of  the  arcs  subtended 
by  the  sides,  another  inscribed  regular 
polygon  is  formed  (having  double  the 
number  of  sides). 

409.  COROLLARY.     If     tangents     are 

drawn  at  the  midpoints  of  the  arcs  between  adjacent  points  of 
contact  of  the  sides  of  a  circumscribed  regular  polygon,  another 
circumscribed  regular  polygon  is  formed  having  double  the 
number  of  sides.  (?). 

410.  COROLLARY.     The  perimeter  of   an  inscribed  regular 
polygon  is  less  than  the  perimeter  of  an  inscribed  regular  poly- 
gon having  twice  as  many  sides,  and  the  perimeter  of  a  cir- 
cumscribed regular  polygon  is  greater  than  the  perimeter  of  a 
circumscribed  regular  polygon  having  twice  as  many  sides. 

(Ax.  12.) 

Ex.  1.  If  in  a  regular  dodecagon  (Fig.  of  403)  all  the  diagonals  are 
drawn  from  vertex  A,  how  many  degrees  are  there  between  adjacent 
pairs  of  diagonals,  at  A  ? 

Ex.  2.  In  the  figure  of  406,  how  many  degrees  are  there  in  each  of 
the  three  angles  at  A  ?  Prove  in  three  ways  that  there  are  120°  in  /.  G. 
If  radii  were  drawn  to  the  vertices  of  the  inscribed  polygon,  what  kind  of 
triangles  would  be  formed  ? 

Ex.  3.  Write  a  formula  for  finding  the  number  of  degrees  in  each 
angle  of  a  regular  polygon. 

Ex.4.  Are  the  following  figures  regular  polygons:  a  square?  an 
isosceles  triangle ?  a  rectangle?  a  rhombus?  an  equilateral  triangle? 

Ex.  6.  In  the  figure  of  406,  prove  that  a  line  from  the  center  of  the 
circle  01  is  the  perpendicular  bisector  of  chord  BC.  Then  prove  tri- 
angles OBI  and  OBM  similar  (M  being  midpoint  of  BC).  Hence  prove 
OI:  OB  =  OB:  OM. 

Ex.  6.  Can  you  explain  how  to  divide  a  circle  into  three  equal  parts  ? 
into  four  equal  parts  ? 


228  BOOK   V.     PLANE   GEOMETRY 

PROPOSITION  III.     THEOREM 

411.   Two  regular  polygons  having  the  same  number  of  sides 
are  similar. 

D' 

Given :  Regular  7i-gons  AD 
and  A'D'. 

To  Prove:  They  are  simi- 
lar. 


Proof: 


Similarly,         £  B  =  Z  £',  Z  C  =  Z  c',  etc. 
That  is,  these  polygons  are  mutually  equiangular. 
Now  AB  =  BC  =  CD  =  etc. ;  A'B'  =  B'  c'  =  C'D'  =  etc.  (402) . 
.  •.  AB  :  A'B'  =  BC  :  B'C'  =  CD:  C'D'  =  etc.       (Ax.  3). 
That  is,  the  homologous  sides  are  proportional. 
Therefore  the  polygons  are  similar  (301). 

Q.E.D. 

PROPOSITION  IV.    THEOREM 

412.   A  circle  can  be  circumscribed  about,  and  a  circle  can 
be  inscribed  in,  any  regular  polygon. 

F^ -^E 

Given :  Regular  polygon  ABCDEF. 

To  Prove:  I.    A   circle  can  be  cir- 
cumscribed about  the  polygon.  Aj^j —  Jj^L.        J^Q 

k. 

\ 
polygon. 


REGULAR  POLYGONS  229 

Proof:  I.    Through  three  consecutive  vertices,  A,  B,  and 

C,  describe  a  circumference,  whose  center  is  O.  Draw  radii 
OA,  OB,  OC,  and  draw  line  OD. 

In  A  AOB  and  COD,               AB  =  CD  (402), 

BO=CO  087). 

Also                                  Z  ABC  ==  Z  BCD  (402), 

Z  OffC  =  Z  ocjg  (55). 

Subtracting,                    Z^J5O  =  ZOCZ)  (Ax.  2). 

.*.  A  AOB  is  congruent  to  A  COD  (52). 

.'.  AO=OD  (27). 

Hence  the  arc  passes  through  D,  and  in  like  manner  it 
may  be  proved  that  it  passes  through  E  and  F. 

That  is,  a  circle  can  be  circumscribed  about  the  polygon. 

II.  AB,  BC,  CD,  DE,  etc.  are  equal  chords  (402). 

.'.  they  are  equally  distant  from  the  center  (208). 

That  is,  a  circle  described,  using  O  as  a  center  and  OM  as 
a  radius,  will  touch  every  side  of  the  polygon.  (202). 

Hence  a  circle  can  be  inscribed  (221).     Q.E.D. 

413.  The  radius  of  a  regular  polygon  is  the  radius  of  the 
circumscribed  circle.     The  radius  of  the  inscribed  circle  is 
called  the  apothem.     The  center  of  a  regular  polygon  is  the 
common  center  of  the  circumscribed  and  inscribed  circles. 

414.  The  central  angle  of  a  regular  polygon  is  the  angle 
included  between  two  radii  drawn  to  the  ends  of  a  side. 

360° 

415.  THEOREM.  Each  central  angle  of  a  regular  n-gon.  = 

Wj 

416.  COROLLARY.     Each  exterior  angle   of  a  regular  n-gon 
_360° 

n 

417.  COROLLARY.    The  radius  drawn  to  any  vertex  of  a 
regular  polygon  bisects  the  angle  at  the  vertex  (95). 

418.  COROLLARY.   The  central   angles  .of  regular  polygons 
having  the  same  number  of  sides  are  equal. 


230 


BOOK   V.     PLANE   GEOMETRY 


PROPOSITION  V.     THEOREM 

419.  The  perimeters  of  two  regular  polygons  having  the 
same  number  of  sides  are  to  each  other  as  their  radii  and  also 
as  their  apothems. 


Given:   Regular  w-gons,  EC,  with  perimeter  P,  radius  E, 
apothem  r ;  and  ErCr  with  perimeter  P',  radius  Rr,  apothem  rf. 


To  Prove :  P  :  P1  =  R  :  Rr  =  r  :  r'. 

Proof:  Draw  radii  OB  and  o1 B1 '. 

In  &AOB  and  AfOfBr,  Z  AOB  =  Z  A'o'B* 

AO  =  BO 
A'O'  =  B'Of 

AO  __  BO 
A'0'~  B'O1 

.*.  A  AOB  is  similar  to  A  A' O'B' 
AB 


A'B' 


But  these  polygons  are  similar 


P_ 
P7 


IL^L 
R'~r' 


AB 


A'B1 
R       r 


(418). 
(187). 

(Ax.  3). 
(306). 
(313). 
(411). 
(317). 

(Ax.  1). 

Q.E.D. 


EEGULAR  POLYGONS 


231 


PROPOSITION  VI.     THEOREM 

420.  The  areas  of  two  regular  polygons  having  the  same 
number  of  sides  are  to  each  other  as  the  squares  of  their  radii 
and  also  as  the  squares  of  their  apothems. 

Given:  Regular  w-gons,  EC  whose  area  is  K,  radius  B, 
apothem  r ;  and  E'C'  whose  area  is  IT',  radius  Rf,  apothem  r1. 

To  Prove :  K :  K'  =  R2 :  R'*  =  r2 :  r'2. 


Proof :  As  in  419, 


Then 


AB  _  R 
~ATB'~~K' 
AB2  E! 


But  these  polygons  are  similar 


AB 


(287). 
(411). 
(376). 

(Ax.  1). 

Q.E.D. 


PROPOSITION  VII.     THEOREM 


421.  The  area  of  a  regular  polygon  is  equal  to  half  the 
product  of  the  perimeter  by  the  apothem.  D 


Given:  (?). 
To  Prove:  (?). 

Proof:  Draw  radii  to  all  the  vertices, 
forming  several  isosceles  triangles. 


Area  of  A  AOB       =  -t 
Area  of  A  BOG       =  < 
Area  of  A  COD       =  < 
etc.,  etc. 

\AB-r 
r  BC-  r 
\CD-r 

Area  of  polygon     =  -, 
Substituting,  area  =  \ 

\(AB  +  J 
^P-  r 

BC+CD 


(364). 

etc.)-r     (Ax.  2). 
(Ax.  6). 

Q.E.D. 


232 


BOOK  V.    PLANE   GEOMETBY 


PROPOSITION  VIII.     THEOREM 

422.  If  the  number  of  sides  of  an  inscribed  regular  polygon 
is  increased  indefinitely,  the  apothem  approaches  the  radius  as 
a  limit. 

Given :  Polygon  FC  inscribed  in  Q  o  ; 
apothem  ='r ;  radius  =  R. 

To  Prove:  That  as  the  number  of 
sides  is  indefinitely  increased,  r  ap- 
proaches R  as  a  limit. 

Proof :  In  the  A  AOK, 

R<r  +  AK  (Ax.  12). 

Or  R-r<AK  (Ax.  7). 

Now,  as  the  number  of  sides  of  the  polygon  is  indefinitely 
increased,  AB  is  indefinitely  decreased. 

Hence  J  AB,  or  AK,  approaches  zero  as  a  limit. 

.*.  R  —  r  approaches  zero  (because  R  —  r<AK). 

That  is,  r  approaches  R  as  a  limit  (227).     Q.E.D. 

NOTE.  It  is  evident  that  if  the  difference  between  two  Variables 
approaches  zero,  either 

(1)  one  is  approaching  the  other  as  a  limit ;  or 

(2)  both  are  approaching  some  third  quantity  as  their  limit. 

423.  THEOREM.     The  circumference  of  a  circle  is  less  than 
the  perimeter  of  any  circumscribed  polygon. 

By  drawing  tangents  at  the  midpoints  of  the  included  arcs 
of  a  circumscribed  polygon,  another  circumscribed  polygon  is 
formed  ;  the  perimeter  of  this  polygon  is  less  than  the  perim- 
eter of  the  given  polygon.  This  can  be  continued  indefinitely, 
decreasing  the  perimeter  of  the  polygons.  Hence  there  can 
be  no  circumscribed  polygon  whose  perimeter  can  be  the  least 
of  all  such  polygons ;  because,  by  increasing  the  number  of 
sides,  the  perimeter  is  lessened.  Hence  the  circumference 
must  be  less  than  the  perimeter  of  any  circumscribed  polygon. 


KEGULAK  POLYGONS  233 

424.  THEOREM.  If  the  number  of  sides  of  an  inscribed 
regular  polygon  and  of  a  circumscribed  regular  polygon  is  in- 
definitely increased. 

I.  The  perimeter  of  each  polygon  approaches  the  circum- 
ference of  the  circle  as  a  limit. 

II.  The  area  of  each  polygon  approaches  the  area  of  the 
circle  as  a  limit. 

Given :  A  circle  O,  whose  circumference 
is  C  and  whose  area  is  5;  AB  and  A'B', 
sides  of  regular  circumscribed  and  in- 
scribed polygons,  having  the  same  number 
of  sides ;  P  and  p',  their  perimeters ;  K 
and  tf,  their  areas. 

To  Prove :  That  if  the  number  of  sides 
is  indefinitely  increased  : 

I.   P  approaches  C  and  Pr  approaches  C  as  limit. 

II.    K  approaches  S  and  E!  approaches  8  as  limit. 

Proof:  I.     The  polygons  are  similar  (431). 


P        OD 

Now,  if  the  number  of  sides  of  these  polygons  is  indefi- 
nitely increased,  OD  approaches  OE  (422). 

X~\  T7T  T> 

Hence    —  approaches  1.     That  is,  —  approaches  1,  or  P 

and  P1  approach  equality  ;   that  is,  they  approach  the  same 
constant  as  a  limit. 

But  P>C  and  C>Pf  and  C  is  constant. 

Hence  P  approaches  C  and  P'  approaches  C.  Q.E.D. 


II.  =  (420). 

If  the  number  of  sides  of  these  polygons  is  indefinitely  in- 

-  o 

creased,  off  approaches  OE2,  and  thus  :  —  -  approaches  unity. 

OD 

(The  argument  continues  the  same  as  in  I.) 
BOBBINS'  s  NEW  PLANE  GEOM.  —  16 


234  BOOK  V.     PLANE   GEOMETRY 

NOTE.  The  theorems  of  423  and  424  are  considered  so  evident,  and 
rigorous  proofs  (as  in  the  case  of  the  demonstrations  for  many  funda- 
mental principles  in  mathematics)  are  so  difficult  for  young  students  to 
comprehend,  that  it  is  advisable  to  omit  the  profound  demonstrations 
and  insert  only  simple  explanations. 

PROPOSITION  IX.    THEOREM 

425.  The  circumferences  of  two  circles  are  to  each  other  as 
their  radii. 

Given :  Two  ©  whose  radii 
are  R  and  Ef  and  circumfer- 
ences, C  and  cf  respectively. 

To  Prove:   C:  Cf  =  R:  it'. 

^    I     >?          ?v       I 

.J 

Proof:   Circumscribe  regular  polygons  (having  the  same 

number  of  sides)  about  these  CD  and  let  P  and  p'  denote 

their  perimeters.     Then  P  :  Pr=R:  Rr  (419). 

Hence  P  •  R!  =  Pr  •  R  (?). 

Now  suppose  the  number  of  sides  of  these  polygons  to  be 

indefinitely  increased, 

P  approaches  C  (424). 

p'  approaches  Cf  (?). 

.*.  P  •  Rf  approaches  C  •  Rr. 
Also  Pr  -  R  approaches  Cf  •  R. 

Hence  C  •  Rf  =  Cr  -  R  (229). 

Therefore  C  :  Cr  =  B  :  Br  (281). 

Q.E.D. 

426.  THEOREM.     The  ratio  of  any  circumference  to  its  diam- 
eter is  constant  for  all  circles.     That  is,  any  circumference 
divided  by  its  diameter  is  the  same  as  any  other  circumference 
divided  by  its  diameter. 

Proof:  =,  (425). 

But 


CIKCLES 


235 


That  is, 


.-.-  =  —, 

D      Dr 
circumference 


=  ft  congtant  for 


(Ax.  1). 

(282). 
Q>E  ^ 


diameter 

427.  Definition  of  IT  (pi).     The  constant  ratio  of  a  circurn- 

y-^ 

ference  to  its  diameter  is  called  TT.     That  is,  —  =  TT. 

D 

The  numerical  value  of  TT  is  3.141592,  or  3^,  approximately. 
(This  is  computed  in  453.) 

428.  FORMULA.     Let  c  =  the  circumference  of  a  circle  with 


radius  1?.     Then  --  =  TT. 


(427). 
(Ax.  3). 


Ex.  1.   Find  the  circumference  of  a  circle  the  radius  of  which  is  12  in. 
Ex.  2.   Find  the  radius  of  a  circle  the  circumference  of  which  is  66  feet. 

Historical  Note.  Gottfried  Wilhelm  von  Leibnitz,  a  German  phi- 
losopher, mathematician,  and  man  of  affairs,  was  born  in  1646  and  died  in 
1716.  He  could  read  Latin  easily  at  12,  and  wrote  some  Latin  verse, 
While  at  the  University  of  Leipsic  he  became  acquainted  with  Francis 
Bacon,  Kepler,  Galileo,  and  Descartes,  modern  thinkers  who  had  revolu- 
tionized science  and  philosophy.  He  resolved  to 
study  mathematics,  but  not  until  he  had  reached  his 
majority  did  he  throw  himself  into  deep  mathemati- 
cal research.  It  was  while  he  was  living  in  Paris  and 
Mainz  that  he  announced  his  imposing  discoveries  in 
natural  philosophy,  mathematics,  mechanics,  optics, 
hydrostatics,  pneumatics,  and  nautical  science.  In 
mathematics,  he  was  the  discoverer  of  the  differ- 
ential and  integral  calculus.  He  possessed  a  marvel-  LEIBNITZ 
ous  ability  for  rapid  and  continuous  work.  Even  in  traveling  his  time 
was  employed  in  solving  mathematical  problems.  He  is  described  as 
moderate  in  habits,  quick  of  temper,  charitable  in  judgment  of  others, 
tolerant  of  differences  of  opinion,  but  impatient  of  contradiction  on 
small  matters  and  desirous  of  honor. 


236  BOOK  V.     PLANE   GEOMETRY 

PROPOSITION  X.     THEOKEM 

429.  The  area  of  a  circle  is  equal  to  half  the  product  of  its 
circumference  by  its  radius. 

.'**•>  P 
Given :  O  with  circumference  C,  area 

-S,  radius  R. 
To  Prove :   s  =  £  c  -  R. 

Proof:  Circumscribe  a  regular  poly- 
gon about  the  circle ;  denote  its  area 
by  JET  and  perimeter  by  P. 

Now  E  =  ±P.R  (421). 

Suppose  the  number  of  sides  of  the  polygon  is  indefinitely 
increased,   .HT  approaches  S,  and  P  approaches  C  (424). 

J  P  •  R  approaches  J  C  •  R  as  a  limit. 

Hence  s  =  ±  c  -  R  (229) .     Q.  E.  D. 

430.  FORMULA.     Let  -S  =  the  area  of  a  circle  whose  circum- 
ference =  C,  and  whose  radius  =  R. 

Then  S=^C-R  (429). 

But  c  =  2irR  (428). 

Substituting,  S  =  irll*  (Ax.  6). 


Ex.  1.   Find  the  area  of  a  circle  the  radius  of  which  is  8  in. 

Ex.  2.   Find  the  radius  of  a  circle  the  area  of  which  is  500  sq.  ft. 

431.  COROLLARY.  The  areas  of  two  circles  are  to  each  other 
as  the  squares  of  their  radii,  and  as  the  squares  of  their  di- 
ameters. 

To  Prove :  s  :  sf  =  E2  :  fi'2  =  D2  :  D'2. 

Proof :  s  =  7TB2,  and  sf  =  TTR'*  (430). 


CIRCLES 


237 


432.  COROLLARY.     The  area  of  a  sector  is  the  same  part  of 
the  circle  as  its  central  angle  is  of  360°.  (Ax.  1.) 

433.  FORMULA.     An  arc  :  circum.  =  central  Z  :  360°  (231). 

.  .  arc  :  2  irJB  =  Z  :  360°. 

NOTE.     If  any  two  of  the  three  quantities,  arc,  R,  Z,  are  known,  the 
remaining  one  can  be  found  by  this  proportion. 

434.  FORMULA.     Sector  :  area  of  O  =  central  Z  :  360° 

(432). 
.-.  sector  :  IT  J£2  =  Z:  360°. 

NOTE.     If  any  two  of  the  three  quantities,  sector,  R,  Z,  are  known, 
the  remaining  one  can  be  found  by  this  proportion. 

435.  FORMULA.     Sector  :  area  of  O  =  arc  :  circum. 

.'.  sector  :  TrR2  =  arc  :  27rR 
.••  sector  =  \  R  •  arc 

436.  Similar  arcs,  similar  sectors,  and 
similar  segments  are  those  which  corre- 
spond to  equal  central  angles,  in  unequal 
circles. 

Thus,  AB,  ArBf,  A!IB"  are  similar  arcs  ; 
AOB,  A'  OB',  and  A"  OB1'  are  similar  sec- 
tors ;  and  the  shaded  segments  are  simi- 
lar segments. 

437.  THEOREM.     Similar  arcs  are  to  each  other  as  their  radii. 
Given:  Arcs  whose  lengths  are  a  and  a',  radii  E  and  Rf  . 
To  Prove  :  a  :  a'  =  E  :  Rr. 

Proof:        • 


(Ax.  1) 


R 


Q.E.D. 


238  BOOK   V.     PLANE   GEOMETRY 

438.    THEOREM.     Similar  sectors  are  to  each  other  as  the 
squares  of  their  radii. 

Given :  Sectors  whose  areas  are  T  and  r',  radii  R  and  JB'.. 
To  Prove :    T  :  Tf  =  R2  :  R'2. 


-y-sS-S  <282> 

Q.E.D. 

PROPOSITION  XI.     THEOREM 

439.   Similar  segments  are  to  each  other  as  the  squares  of 
their  radii.  o' 

Given :   Similar  segments  ABC 
and  A'B'C'. 


ent  A'B'C'  =  R2:  Rf2.                            ^^          ftT^m 
Proof  :   A  AOB  and  A'O'B'  are  similar 
A  AOB        R2 

(306). 
(375). 

Ax.  1). 

(282). 
f285\ 

Also 
sector 

sector  OACB         R2 

sector  O'A'C'B'     R'2 

sector  CMC£           A  AOB                            ^ 

'  sector  O'A'C'B'  ~  A  A'O'B' 
sector  OACB      sector  O'A'C'B' 

AAOB             A  A'  O'B' 

OACB  —  A  AOB      sector  O'A'C'B'  —  A  ^'o'j?' 

That  is          segment  ABC  _  segment  A'B'C' 

AAOB  A  A' O'B' 

segment  ABC  _   AAOB  _  R2 
*  segment  A'B'C'  ~~  A  A'O'B'  ~~  R'2 


ORIGINAL   EXERCISES  239 

ORIGINAL  EXERCISES    (THEOREMS) 

1.  The  central  angle  of  a  regular  polygon  is  the  supplement  of  the 
angle  of  the  polygon. 

2.  An  equiangular  polygon  inscribed  in  a  circle  is  regular  (if  the 
number  of  its  sides  is  odd). 

3.  An  equiangular  polygon  circumscribed  about  a  circle  is  regular. 
[Draw  radii  and  apothems.] 

4.  The  sides  of  a  circumscribed  regular  polygon  are  bisected  at  the 
points  of  contact. 

5.  The  diagonals  of  a  regular  pentagon  are  equal. 

6.  The  diagonals  drawn  from  any  vertex  of  a  regular  n-gon  divide 
the  angle  at  that  vertex  into  n  —  2  equal  parts. 

7.  If  a  regular  polygon  is  inscribed  in  a  circle,  and  another  regular 
polygon  having  the  same  number  of  sides  is  circumscribed  about  it,  the 
radius  of  the  circle  is  a  mean  proportional  between  the  apothem  of  the 
inner  and  the  radius  of  the  outer  polygon. 

8.  The  area  of  the  square  inscribed  in  a  sector 
the  central  angle  of  which  is  a  right  angle,  is  equal  to 
half  the  square  of  the  radius. 

[Find  x2,  the  area  of  OEDC.~] 

9.  The  apothem  of  an  equilateral  triangle  is  one 
third  the  altitude  of  the  triangle. 

10.  The  chord  that  bisects  a  radius  of  a  circle  at  right  angles  is  the 
side  of  the  inscribed  equilateral  triangle. 

[Prove  that  the  central  Z  subtended  is  120°.] 

11.  If  ABODE  is   a  regular  pentagon,  and 
diagonals  A  C  and  BD  are  drawn,  meeting  at  0  : 

(a)  A0  =  AB. 
(l>)  AO  is  II  to  ED. 

(c)  A  BOC  is  similar  to  A  BDC. 

(d)  ZACB  =  36°. 

(e)  AC  is   divided   into   mean  and  extreme 
ratio  at  O. 

12.  The  altitude  of  an  equilateral  triangle  is  three  fourths  the  diameter 
of  the  circumscribed  circle. 

13.  The  apothem  of  an  inscribed  regular  hexagon  equals  half  the  side 
of  an  inscribed  equilateral  triangle. 


240 


BOOK   V.     PLANE   GEOMETRY 


14.  The  area  of  a  circle  is  four  times  the   area  of   another  circle 
described  upon  its  radius  as  a  diameter. 

15.  The  area  of  an  inscribed  square  is  half  the  area  of  the  circum- 
scribed square. 

16.  An  equilateral  polygon  circumscribed   about  a  circle  is  regular 
(if  the  number  of  its  sides  is  odd) . 

17.  The  sum  of  the  circles  described  upon  the  legs  of  a  right  triangle 
as  diameters  is  equal  to  the  circle  described  upon 

the  hypotenuse  as  a  diameter. 

18.  A  circular  ring  (the  area  between  two  con- 
centric circles)  is  equal  to  the  circle  described  upon 
the  chord  of  the  larger  circle,  which  is  tangent  to 
the  less,  as  a  diameter. 

Proof :  Draw  radiiOB,  OC.     A  OBC  is  rt.  A  (?)  ; 
and  OC2  -  OB2  =  BC2  (?).     Etc. 

19.  If  semicircles  are  described  upon  the  three 
sides  of  a  right  triangle  (on  the  same  side  of  the 
hypotenuse),  the  sum  of  the  two  crescents  thus 
formed  is  equal  to  the  area  of  the  triangle. 


Proof: 


Entire  figure  =  \wAJi   +  crescent  BDC  +  crescent  A  EC  (?) 
Entire  figure  =  £  irAC2  +  |  TrBC2  +  A  AB C  (?) 


Now  use  Ax.  1  ;  etc. 

20.  Show  that  the  theorem  of  Ex.  19  is  true  in  the  case  of  a  right 
triangle  whose  legs  are  18  and  24. 

21.  If  from  any  point  in  a  semicircle  a  line  is  drawn  perpendicular 
to  the  diameter  and  if  semicircles  are  described  „ 

on  the  two  segments  of  the  original  diameter 
as  diameters,  the  area  of  the  surface  bounded  by 
these  three  semicircles  equals  the  area  of  a  circle 
whose  diameter  is  the  perpendicular  first  drawn. 


Proof: 


-  J,r(f)2-i,r(f)2  = 


22.    Show  that  the  theorem  of  Ex.  21"  is  true  in  the  case  of  a  circle 
with  diameter  AB  equal  to  25  and  AD  equal  to  5. 


ORIGINAL   EXERCISES  241 

23.  If  the  sides  of  a  circumscribed  regular  polygon  are  tangent  to  the 
circle  at  the  vertices  of  an  inscribed  regular  polygon,  each  vertex  of  the 
outer  lies  on  the  prolongation  of   the  apothems  of  the  inner  polygon, 
drawn  perpendicular  to  the  several  sides. 

24.  The  sum  of  the  perpendiculars  drawn  from  any  point  within  a 
regular  n-gon  to  the  several  sides  is  constant  [=  n  •  apothem]. 

25.  The  area  of  a  circumscribed  equilateral  triangle  is  four  times  the 
area  of  the  inscribed  equilateral  triangle. 

26.  If  a  point  is  taken  dividing  the  diameter  of  a  circle  into  two 
parts  and  circles  are  described  upon  these  parts  as  diameters,  the  sum  of 
the  circumferences  of  these  two  circles  equals  the  circumference  of  the 
original  circle. 

27.  Show  that  the  theorem  of  Ex.  26  is  true  in  the  case  of  a  circle 
the  segments  of  the  diameter  of  which  are  7  and  12. 

28.  The  area  of  an  inscribed  regular  octagon  is  equal  to  the  product 
of  the  diameter  by  the  side  of  the  inscribed  square. 

29.  If  squares  are  described  on  the  six  sides  of  a  regular  hexagon 
(externally),  the  twelve  exterior  vertices  of  these  squares  are  the  vertices 
of  a  regular  12-gon. 

30.  If  the  alternate  vertices  of  a  regular  hexagon  are  joined  by  draw- 
ing diagonals^  another  regular  hexagon  is   formed.     Also  its  area  is  one 
third  of  the  original  hexagon. 

31.  Show  that  the  theorem  of  Ex.  18  is  true  in  the  case  of  two  con- 
centric circles  whose  radii  are  34  and  16. 

32.  In  the  same  or  equal  circles  two  sectors  are  to  each  other  as  their 
central  angles. 

33.  If  the  diameter  of  a  circle  is  10  in.  and  a  point  is  taken  dividing 
the  diameter  into  segments  with  lengths  4  in.  and  6  in.,  and  on  these 
segments  as  diameters  semicircles  are  described  on  opposite  sides  of  the 
diameter,  these  arcs  form  a  curved  line  which  divides  the  original  circle 
into  two  parts  in  the  ratio  of  2  :  3. 

34.  If  the  diameter  of  a  circle  is  d  and  a  point  is  taken  dividing  the 
diameter  into  segments  with  lengths  a  and  d  —  a,  and  on  these  segments 
as  diameters  semicircles  are  described  on  opposite  sides  of  the  diameter, 
these  arcs  form  a  curved  line  which  divides  the  original  circle  into  two 
parts  in  the  ratio  of  a  :  d  —  a. 


242 


BOOK   V.     PLANE   GEOMETRY 


CONSTRUCTION  PROBLEMS 

PROPOSITION  XII.     PROBLEM 
440.  To  inscribe  a  square  in  a  given  circle. 
Given:  The  circle  o. 
Required :  To  inscribe  a  square. 
Construction :  Draw  any  diameter,  AB, 
and   another   diameter,    CD,   JL   to   AB. 
Draw  AC,  BC,  BD,  AD. 

Statement:    ACBD    is    an     inscribed 
square.  Q.E.F. 

Proof :  The  central  A  are  all  equal 

.-.  arcs  AC,  CB,  BD,  DA  are  equal 
.•.  ACBD  is  an  inscribed  square 


(42). 
(193). 

(404).       Q.B.D. 


PROPOSITION  XIII.     PROBLEM 
441.   To  inscribe  a  regular  hexagon  in  a  given  circle. 
Given:  (?).  Required:  (?). 

Construction:  Draw  any  radius,  A  O. 
At  A,  with  radius  =  to  AO,  describe  arc 
intersecting  the  given  O  at  B.  Draw 
AB. 

Statement:  AB  is  the  side  of  an  in- 
scribed regular  hexagon. 

Proof:  Draw  BO.     A  ABO  is  equilateral  (Const.). 

.•.  A  ABO  is  equiangular  (56). 

.-.  Z  AOB  =  60°  (109). 

That  is,  arc  AB  =  i  of  the  circumference  ;  and  if  arc  AB 
is  used  as  a  unit,  it  divides  the  circumference  into  6  equal 
arcs.  If  the  chords  are  drawn,  an  inscribed  regular  hexa- 
gon is  formed  (404). 

Q.E.D. 


CONSTRUCTION   PROBLEMS 


243 


PROPOSITION  XIV.     PROBLEM 

442.    To  inscribe  a  regular  decagon  in  a  given  circle. 
Given:   (?). 
Required:    (?). 

Construction:  Draw  any  radius  AO. 
Divide  it  into  mean  and  extreme  ratio 
(by  349),  having  the  larger  segment  next 
to  the  center.  Taking  A  as  a  center  and  OB 
as  a  radius,  draw  an  arc  cutting  O  at  c. 
Draw  AC,  EC,  OC. 

Statement  :  AC  is  a  side  of  the  inscribed 
regular  decagon. 

Proof:  AO:BO  =  BO:AB 

Substituting,  AO  :  AC  =  AC:  AB 

.-.  A  ABC  and  AOC  are  similar 
.-.  First,  Z.ACB  =  ZO 
Second,  A  ABC  is  isosceles 
.'.  AC—  BC 

But  AC=BO 

.'.  BC  =  BO 

.'.  z.  BCO  =  Z  o 

Now  Z  AGO  ==  2  Z  O 


Q.E.F. 

(Const.). 
(Ax.  6). 
(306). 
(312). 

(similar  to  A  AOC). 
(24). 
(Const.). 
(Ax.  1). 
(55). 


And 
Adding, 


(Ax.  4). 
(55). 


Z  o  =  1  Z  o 


the  A  of  &ACO^~5ZTd  (Ax.  2). 

/.  5Z  0  =  180°  (104). 

.'.  Z  o  =  36°  =  iV  of  360°  (Ax.  3). 

.'.  arc  AC=  -^Q  of  the  circumference 

(193). 

That  is,  if  arc  AC  is  used  as  a  unit  it  divides  the  circum- 
ference into  ten  equal  arcs  ;  and  if  the  chords  are  drawn,  an 
inscribed  regular  decagon  is  formed.  (404). 

Q.E.P. 


244  BOOK  V.     PLANE   GEOMETRY 

PROPOSITION  XV.     PROBLEM 

443.  To  inscribe  a  regular  i5-gon  (pentadecagon)  in  a  given 
circle. 

Given:    (?).  Required:    (?). 

Construction:  Draw  AB,  the  side  of  an 
inscribed  hexagon,  and  AC,  the  side  of  an 
inscribed  decagon.  Draw  BC. 

Statement :  BC  is  the  side  of  an  inscribed 
regular  15-gon.  Q.E.F. 

Proof :          Arc  BC  =  arc  AB  —  arc  AC 

=  1—  y1^-  of  circumference       (Const.). 
=  ^  of  circumference. 

That  is,  if  arc  BC  is  used  as  a  unit,  it  divides  the  circum- 
ference into  fifteen  equal  arcs;  and  if  the  chords  are  drawn, 
an  inscribed  regular  pentadecagon  is  formed  (404).  Q.E.D. 

444.  To  inscribe  in  a  given  circle: 

I.    A  regular  8-gon,  a  regular  i6-gon,  a  regular  32-gon,  etc. 

II.    A  regular  i2-gon,  24-gon,  etc. 
III.    A  regular  ao-gon,  6o-gon,  etc. 

Construction :   I.    Inscribe  a  square.  (440.) 

Bisect  the  arcs  and  draw  the  chords.     Proof  :  (404). 

II.    Inscribe  a  regular  hexagon,  etc. 
III.    Inscribe  a  regular  pentadecagon,  etc. 

445.  To  inscribe  an  equilateral  triangle  in  a  circle. 
Construction:   Join  the  alternate  vertices  of  an  inscribed 

regular  hexagon.     Proof :   (?)  (405). 

446.  To  inscribe  a  regular  pentagon  in  a  given  circle. 

447.  To  circumscribe  a  regular  polygon  about  a  circle. 
Construction :   Inscribe  a  polygon  having  the  same  number 

of  sides.     At  the  several  vertices  draw  tangents. 

Statement:   (?).     Proof:   (?)  (406). 


CONSTRUCTION   PROBLEMS  245 


FORMULAS 


448.    Sides  of  inscribed  polygons. 
1.    The  side  of  inscribed  equilateral 
triangle  =  B  V3. 


Proof:  Z  ACB  is  a  rt.  Z  (240). 

AB  =  2R  (189). 

BC=R  (441). 

z2=(2B)2-E2  (335). 

x=  R  V3.  Q.E.D. 

2.  The  side  of  an  inscribed  square  =  z?V2. 
Proof :  In  fig.  of  440 

Ic2  =  B24-«2  (334). 

.'.  AC=RV2.  Q.E.D. 

3.  The  side  of  an  inscribed  regular  hexagon  =  R        (441). 

4.  The  side  of  an  inscribed  regular  decagon  =  |  R  ( V5  —  1) 

(352;  442). 

D  B 

449.  Sides  of  circumscribed  poly- 
gons. 

1.  The  side  of  a  circumscribed 
equilateral  A  =  2  R  V3. 


Proof:  ^DAB  =  Z  DBA  =  ^D=60°        (?). 
.-.  A  ABD  is  equilateral. 

AD^AB^B^  (448). 

.'.  DF=  2BV3.  Q.E.D. 

2.   The  side  of  a  circumscribed  square  ~  2  R  (?)• 


246 


BOOK  V.    PLANE   GEOMETRY 


3.   The  side  of  a  circumscribed  regular  hexagon  =  f  R  V3. 

Proof :  This  side  =  a  side  of  an  equilateral  A  with  altitude 
R.     Let  x  =  this  side. 

450.  In  an  equilateral  triangle,  apothem  =  J  R. 
Proof :  Fig.  of  449.     Bisect  arc  AC  at  H. 
Draw  OA,  OC,  AH  and  CH. 

Figure  AOCH  is  a  rhombus  (448,  3). 

.•.Otf  =  *OH=iJS  ^   (135). 

Q.E.D. 

PROPOSITION  XVI.     PROBLEM 

451.  In  a  circle  whose  radius  is  R  is  inscribed  a  regular 
polygon  whose  side  is  s ;  to  find  the  formula  for  the  side  of  an 
inscribed  regular  polygon  having  double  the  number  of  sides. 

Given :  AB  =  s,  a  side  of  an  inscribed 
regular  polygon  in  O  whose  radius  is  R ; 
C,  the  midpoint  of  arc  AB ;  chord  AC. 

Required:  To  find  the  value  of  AC, 
the  side  of  a  regular  polygon  having 
double  the  number  of  sides  and  inscribed 
in  the  same  circle. 

Construction :  Draw  radii  OA  and  OC. 

Computation :  OC  bisects  AB  at  right  A 

In  rt.  A  AON,  O  is  an  acute  Z.     Hence  in  A  AOC, 

A(?=~OA2+W?-2>OC'  ON  (337). 


(83). 


But 
And 

Substituting, 

AO  =  OC  =  R 

(187). 
(335). 
(Ax.  6). 

Q.E.F. 

ON=  Vfl2-(is)2  = 
Ic2=2fl2-2JR.  JV< 

|V4  E2  —  s2 

±*2-*2 

.-.  AC  =  \/2  R2  -  R  V4  R' 

J_  j2p 

452.    FORMULA.     If  R  =  1,  and  given  side  =  s,  the  side  of  a 
regular  polygon  having  twice  as  many  sides  =  \/2—  V4—  s2. 


CONSTRUCTION   PROBLEMS  247 

PROPOSITION  XVII.     PROBLEM 
453.  To  find  the  approximate  numerical  value  of  IT, 
Given :  A  circle  whose  diameter  =  D  ;  circumference  =  C. 

Required :  The  value  of  TT,  that  is  — . 

Method:  1.   We  may  select  a  O  of  any  diameter.      (426.) 

2.  We  can  compute  the  side  and  the  perimeter  of  some  in- 
scribed regular  polygon.  (448.) 

3.  We  can  compute  the  side  and   perimeter  of   another 
inscribed   regular    polygon   having   double   the   number  of 
sides.  (451.) 

4.  We  can  now  compute  the  side  and  perimeter  of  a  third 
inscribed  regular  polygon,  having  still  double  the  number  of 
sides.  (451.) 

5.  By  continuing  this  process  until  the  consecutive  perim- 
eters differ  very  slightly,  we  can  find  the  approximate  value 

of  the  circumference. 

/•* 

6.  Thus,  knowing  both  C  and  D,  we  know  —  or  TT. 

Computation :  1.    For  simplicity,  take  D  =  2,  and  R  =  1. 

2.  We  will  select  the  regular  hexagon  as  the  first  polygon. 

.*.  S6  =  1,  and  P6  =  6. 

3.  .'.  512  =  V2  _  V4  -  S62  =  V2  _  V3  =  .5176381    (452). 
and  P12  =  6.2116572. 

4.  Hence  S24  =  ^2  _  V4  - S122  =  V2  _  V4^(751763S1)2 

=  .2610524. 
Also  P24  =  6.2652576. 

5.  By  continuing,  S3072  =  .002045. 
Also  P3072  =  6.283184. 

6.  It  now  appears  that,  approximately,  C  =  6.283184. 

Hence  TT  =  6'288184  =  3.141592+.       Q.  E.F. 


248  BOOK   V.     PLANE   GEOMETRY 

This  calculation  is  tabulated  for  reference. 

*6  =  1,  .-.  P6  =  6.  Sw2  =  0.032723,  .-.  P192  =  6.282904. 

sn  =  0.517638,  .-.  P12  =  6.211657.  ,«384  =  0.016362,  .-.  P884  =  6.283115. 
*24  =  0.261052,  .-.  P24  =  6.265257.  sm  =  0.008181,  .-.  P768  =  6.283169. 
S48  =  0.130806,  .-.  P48  =  6.278700.  »15,6  =  0.004091,  .-.  P1536  =  6.283180. 
s96  =  0.065438,  .-.  P96  =  6.282053.  s3072  =  0.002045,  .-.  P8072  =  6.283184. 

ORIGINAL  EXERCISES    (NUMERICAL) 
MENSURATION  OF  REGULAR  POLYGONS  AND  THE  CIRCLE 

1.  Find  the  angle  and  the  central  angle  of  : 

(£)  a  regular  pentagon  ;    (ii)  a  regular  octagon ;    (Hi)  a  regular  do- 
decagon ;  (iv)  a  regular  20-gon. 

2.  Find  the  area  of  a  regular  hexagon  whose  side  is  8. 

3.  Find  the  area  of  a  regular  hexagon  whose  apothem  is  4. 

4.  In  a  circle  whose  radius  is  10  are  inscribed  an  equilateral  triangle, 
a  square,  and  a  regular  hexagon.     Find  the  perimeter,  the  apothem,  and 
the  area  of  each. 

6.  About  a  circle  whose  radius  is  10  are  circumscribed  an  equilateral 
triangle,  a  square,  and  a  regular  hexagon.  Find  the  perimeter  and  the 
area  of  each. 

6.  Find  the  circumference  and  the  area  of  a  circle  whose  radius  is  5 
inches.     [Use  *  =  3}.] 

7.  Find  the  circumference  and  the  area  of  a  circle  whose  diameter  is 
42  centimeters. 

8.  The  radius  of  a  certain  circle  is  9  meters.     What  is  the  radius  of 
a  second  circle  whose  circumference  is  twice  as  long  as  the  first  ?  of  a 
third  circle  whose  area  is  twice  as  great  as  the  first  ? 

9.  If  the  circumference  of  a  circle  is  55  yards,  what  is  its  diameter  ? 

10.  If  the  area  of  a  circle  is  113f  square  meters,  what  is  its  radius  ? 

11.  In  a  circle  whose  radius  is  35  there  is  a  sector  whose  angle  is  40°. 
Find  the  length  of  the  arc  and  the  area  of  the  sector. 

12.  The  area  of  a  circle  is  6£  times  the  area  of  another.     If  the  radius 
of  the  smaller  circle  is  12,  what  is  the  radius  of  the  larger  circle? 

13.  If  the  angle  of  a  sector  is  72°  and  its  arc  is  44  inches,  what  is  the 
radius  of  the  circle?    What  is  the  area  of  the  sector? 

14.  In  a  circle  whose  radius  is  7  find  the  area  of  the  segment  whose 
central  angle  is  120°.     [Required  area  =  1  (area  of  O  —  area  eq.  A.] 


ORIGINAL  EXERCISES  249 

15.  If  the  radius  of  a  circle  is  4  feet,  what  is  the  area  of  a  segment 
whose  arc  is  60°?  of  a  segment  whose  arc  is  a  quadrant? 

16.  Find  the  area  of  a  circle  inscribed  in  a  square  whose  area  is  75. 

17.  Find  the  area  of  an  equilateral  triangle  inscribed  in  a  circle  whose 
area  is  441  TT  square  meters. 

18.  If  the  length  of  a  quadrant  is  8  inches,  what  is  the  radius  ? 

19.  Find  the  length  of  an  arc  subtended  by  the  side  of  an  inscribed 
regular  15-gon  if  the  radius  is  4|  inches. 

20.  The  side  of  an  equilateral  triangle  is  10.    Find  the  areas  of  its  in- 
scribed and  circumscribed  circles. 

21.  Find  the  perimeter  and  the  area  of  a  segment  whose  chord  is  the 
side  of  an  inscribed  regular  hexagon,  if  the  radius  of  a  circle  is  5^. 

22.  A  circular  lake  9  rods  in  diameter  is  surrounded  by  a  walk  \  rod 
wide.     What  is  the  area  of  the  walk  ? 

23.  A  locomotive  driving  wheel  is  7  feet  in  diameter.     How  many 
revolutions  will  it  make  in  running  a  mile  ? 

24.  What  is  the  number  of  degrees  in  the  central  angle  whose  arc  is 
as  long  as  the  radius  ? 

25.  Find  the  side  of  the  square  equal  to  a  circle  whose  diameter  is 
4.2  meters. 

26.  Find  the  radius  of  that  circle  equal  to  a  square  whose  side  is  5.5 
inches. 

27.  Find  the  radius  of  the  circle  which  divides  a  given  circle  whose 
radius  is  10£  into  two  equal  parts. 

28.  Three  equal  circles  are  each  tangent  to  the  other  two  and  the 
diameter  of  each  is  40  feet.     Find  the  area  between  these  circles. 

[Required  area  =  area  of  an  eq.  A  minus  area  of  three  sectors.] 

29.  Find  the  area  of  the  three  segments  of  a  circle  whose  radius  is 
5\/3,  formed  by  the  sides  of  the  inscribed  equilateral  triangle. 

30.  If  a  cistern  can  be  emptied  in  5  hours  by  a  2-inch  pipe,  how  long 
will  be  required  to  empty  it  by  a  1-inch  pipe  ? 

31.  Find  the  side,  the  apothem,  and  the  area  of  a  regular  decagon  in- 
scribed in  a  circle  whose  radius  is  6  feet. 

32.  What  is  the  area  of  the  circle  circumscribed  about  an  equilateral 
triangle  whose  area  is  48  V3  ? 

33.  The  circumferences  of  two  concentric  circles  are  40  inches  and  50 
inches.     Find  the  area  of  the  circular  ring  between  them. 

ROBBINS'S   NEW    PLANE    GEOM.  —  17 


250  BOOK   V.     PLANE   GEOMETRY 

34.   A  circle  has  an  area  of  80  square  feet.     Find  the  length  of  an  arc 
of  80°. 

36.    Find  the  angle  of  a  sector  whose  perimeter  equals  the  circum- 
ference. 

36.  Find  the  angle  of  a  sector  whose  area  is  equal  to  the  square  of 
the  radius. 

37.  Find  the  area  of  a  regular  octagon  inscribed  in  a  circle  whose 
radius  is  20. 

[Inscribe  square,  then  octagon.     Draw  radii  of  octagon.     Find  area 
of  one  isosceles  A  formed,  whose  altitude  is  half  the  side  of  the  square.] 

38.  A  rectangle  whose  length  is  double  its  width,  a  square,  an  equi- 
lateral triangle,  and  a  circle  all  have  the  same  perimeter,  namely,  132 
meters.     Which  has  the  greatest  area?  the  least? 

39.  Through  a  point  without  a  circle  whose  radius  is  35  inches  two 
tangents  are  drawn,  forming  an  angle  of  60°.     Find  the  perimeter  and  the 
area  of  the  figure  bounded  by  the  tangents  and  their  smaller  intercepted 
arc. 

40.  In  a  circle  whose  radius  is  12  are  two  parallel  chords  which  sub- 
tend arcs  of  60°  and  90°  respectively.     Find  the  perimeter  and  the  area 
of  the  figure  bounded  by  these  chords  and  their  intercepted  arcs. 

41.  A  quarter  mile  race  track  is  to  be  laid  out,  having  parallel  sides 
but  semicircular  ends  with  radius  105  feet.      Find  the  length  of  the 
parallel  sides. 

42.  If  the  diameter  of  the  earth  is  7920  miles,  how  far  at  sea  can  the 
light  from  a  lighthouse  150  feet  high  be  seen  ? 

43.  The  diameter  of  a  circle  is  18  inches.     Find  the  area  of  the  figure 
between  this  circle  and  the  circumscribed  equi- 
lateral triangle. 

44.  How  far  does  the  end  of  the  minute  hand 
of  a  clock  move  in  20  minutes,  if  the  hand  is  3£ 
inches  long  ? 

46.  The  diameter  of  a  circle  is  16  inches. 
What  is  the  area  of  that  portion  of  the  circle 
outside  the  inscribed  regular  hexagon? 

46.   Using  the  vertices  of  a  square  whose  side 

is  12,  as  centers,  and  radii  equal  to  4,  four  quadrants  are  described  within 
the  square.     Find  the  perimeter  and  the  area  of  the  figure  thus  formed. 


ORIGINAL   EXERCISES 


251 


47.  Using  the  four  vertices  of  a  square  whose 
side  is  12  as  centers,  and   radii   equal  to  6,  four 
arcs  are  described  without  the  square  (see  figure). 
Find  the  perimeter  and  the  area  of  the  figure  bounded 
by  these  four  arcs. 

48.  Using  the  vertices  of  an  equilateral  triangle 
whose  side  is  16  as  centers  and  radii  equal  to  8, 
three  arcs  are  described  within  the  triangle.     Find 

the  perimeter  and  the  area  of  the  figure  bounded  by  these  arcs.     Do  the 
same  if  the  three  arcs  are  described  without  the  triangle. 

49.  Using  the  vertices  of   a  regular   hexagon,  whose   side  is  20,  as 
centers  and  radii  equal  to  10,  six  arcs  are  described  within  the  hexagon. 
Find  the  perimeter  and  the  area  of  the  figure  bounded  by  these  arcs.     Do 
the  same  if  the  six  are  described  without  the  hexagon. 

60.  If  semicircumferences  are  described  with- 
in  a  square,  with  side  8  inches,  upon  the  four 
sides  as  diameters,  find  the  areas   of  the   four 
lobes  bounded  by  the  eight  quadrants.     Find 
the  area  of  any  one. 

In  the  following  exercises  let  n  =  number  of 
sides  of  the  regular  polygon  ;  s  =  length  of  side  ; 
r  =  apothem  ;  R  =  radius  ;  K  =  area. 

61.  If  n  =  3,  show  that  s  =  U  V3  ;  r  =  \  #  ;  K  =  3  R*^  =  3  r2V3. 

4 

62.  If  n  =  4,  show  that  s  =  R\/2  =  2r;  K  =  2  R2  =  4  r2. 


53.   If  »  =  6,  show  that,  =  R=-;  K  = 

3 


=2r* 


64.   If  n  =  8,  show  that  s  =  R  \/2  -  v/2  =  2  r(  V2-  1)  ;  r=  —  V2+\/2~; 


65.  If  n  =  10,  show  that  s=~  (  V5  -  1)  ;  r  =  —  VlO  +  2V5. 

66.  Ifn  =    5,  show  that  s=—  VlO  -2V5;  r  =—  (V5  +  1). 
57.    If  n  =  12,  show  that  s  =  R~Vz  -  V3  =  2  r  (2  -  V3)  ; 

72  =  2rV2-V3i;    r=  -»  V2+V3;  K  =  12  r2(2  -  V3)  = 


252  BOOK   V.    PLANE   GEOMETRY 

58.  The  apothem  of  a  regular  hexagon  is  18  V3  inches.  Find  its 
side  and  area.  Find  the  area  of  the  circle  circumscribed  about  it. 

69.  What  is  the  radius  of  a  circle  whose  area  is  doubled  by  increas- 
ing the  radius  10  feet? 

60.  If  an  8-inch  pipe  will  fill  a  cistern  in  3  hours  20  minutes,  how 
long  will  it  require  a  2-inch  pipe  to  fill  it  ? 

61.  The  radius  of  a  circle  is  12  meters.     Find : 
(a)    The  area  of  the  inscribed  square. 

(&)  The  area  of  the  inscribed  equilateral  triangle. 

(c)  The  area  of  the  inscribed  regular  hexagon. 

(d)  The  area  of  the  inscribed  regular  dodecagon. 

(e)  The  area  of  the  circumscribed  square. 

(/)  The  area  of  the  circumscribed  equilateral  triangle. 
(<?)    The  area  of  the  circumscribed  regular  hexagon. 
(h)    The  area  of  the  circumscribed  regular  dodecagon. 

62.  The  radius  of  a  circle  is  18.     Find : 

(a)    The  side  and  the  apothem  of  the  inscribed  square. 

(6)     The  side  and  the  apothem  of  the  inscribed  equilateral  triangle. 

(c)     The  side  and  the  apothem  of  the  inscribed  regular  hexagon. 

(rf)    The  area  of  the  inscribed  square. 

(e)    The  area  of  the  inscribed  equilateral  triangle. 

(/)  The  area  of  the  inscribed  regular  hexagon. 

(</)    The  area  of  the  inscribed  regular  octagon. 

(k)    The  area  of  the  circumscribed  regular  hexagon. 

63.  Prove  that  the  area  of  an  inscribed  regular  hexagon  is  a  mean 
proportional  between  the  areas  of  the  inscribed  and  the  circumscribed 
equilateral  triangles.     [Find  the  three  areas  in  terms  of  R.~] 

64.  AB  is  one  side  of  an  inscribed  equilateral  triangle,  and  C  is  the 
midpoint  of  AB.     If  AB  is  prolonged  to  O  making  BO  equal  to  BC, 
and  OT  is  drawn  tangent  to  the  circle  at  T,  OT  is  f  the  radius. 

66.  A  square,  an  equilateral  triangle,  a  regular  hexagon,  and  a  circle 
all  have  the  same  area,  namely  5544  sq.  ft.  Which  figure  has  the  least 
perimeter  ?  the  greatest  ? 

66.  A  square,  an  equilateral  triangle,  a  regular  hexagon,  and  a  circle 
all   have  the  same   perimeter,  namely  396   in.     Find  their  areas   and 
compare  them. 

67.  The  circumferences  of  two  concentric  circles  are  330  and  440  in. 
respectively.     Find  the  radius  of  another  circle  equivalent  to  the  ring 
between  these  two  circles. 


ORIGINAL  CONSTRUCTIONS 


253 


ORIGINAL  CONSTRUCTIONS 

It  is  required : 

1.  To  circumscribe  a  regular  hexagon  about  a  given  circle. 

2.  To  circumscribe  an  equilateral  triangle  about  a  given  circle. 

3.  To  circumscribe  a  regular  decagon  about  a  given  circle ;  a  regular 
16-gon  ;  a  regular  24-gon ;  a  square. 

4.  To  construct  an  angle  of  36°;  of  18°;  of  72°;  of  24°;  of  6°;  of 
48°;  of  96°. 

5.  To  construct  a  regular  hexagon  upon  a  given  line  as  a  side. 

6.  To  construct  a  regular  decagon  upon  a  given  line  as  a  side. 

7.  To  construct  a  regular  octagon  upon  a  given  line  as  a  side. 

8.  To  construct  a  regular  pentagon  upon  a  given  line  as  a  side. 

9.  To  construct  a  square  with  double  the  area  of  a  given  square. 

10.  To  inscribe  in  a  given 
circle  a  regular  polygon  simi- 
lar to  a  given  regular  polygon. 

Construction :  From  the 
center  of  the  polygon  draw 
radii.  At  the  center  of  the 
circle  construct  A  =  to  these 
central  A  orf  the  polygon 
Draw  chords.  Etc. 

11.  To  construct  a  regular  pentagon  which  shall  have  double  the  area 
of  a  given  regular  pentagon. 

12.  To  construct  a  circumference  equal  to  the  sum  of  two  given  cir- 
cumferences. 

13.  To  construct  a  circumference  which  shall  be  three  times  a  given 
circumference. 

14.  To  construct  a  circumference  equal  to  the  difference  of  two  given 
circumferences. 

16.   To  construct  a  circle  whose  area  shall  be  five  times  a  given  circle. 

16.  To  construct  a  circle  equal  to  the    sum  of  two  given  circles ; 
another,  equal  to  their  difference. 

17.  To  construct  a  circle  whose  area  shall  be  half  a  given  circle. 

18.  To  bisect  the  area  of  a  given  circle  by  a  concentric  circle. 

19.  To  divide  a  given  circumference  into  two  parts  which  shall  be  in 
the  ratio  of  3  :  7  ;  into  two  other  parts  which  shall  be  in  the  ratio  of  5  :  7 ; 
into  still  two  other  parts,  in  the  ratio  of  8  :  7. 


254 


BOOK  V.     PLANE   GEOMETRY 


MAXIMA  AND   MINIMA 

454.  Of  geometrical  magnitudes  that  satisfy  a  given  con- 
dition (or  given  conditions)  the  greatest  is  called  the  maxi- 
mum, and  the  least,  the  minimum. 

Thus,  of  all  chords  that  can  be  drawn  through  a  given  point  within 
a  circle,  the  diameter  is  the  maximum,  and  the  chord  perpendicular  to 
the  diameter  at  the  point  is  the  minimum. 

Isoperimetric  figures  are  figures  having  equal  perimeters. 

PROPOSITION  XVIII.     THEOREM 

455.  Of  all  triangles  having  two  given  sides,  that  imwhich 
these  sides  form  a  right  angle  is  the  maximum. 

Given :  A  ABC  and  A  ABD 
having  AB  common,  and 
.4(7  =  to  AD',  Z  CAB  a  rt.  Z 
and  Z  DAB  not  a  right  Z. 

To  Prove:  AABC>AABD. 

Proof :   Draw  altitude  DE. 

Now  AD>DE 

.'.  AODE 

Multiply  each  member  by  J  AB. 


(87). 
(Ax.  6). 


Then  |  AB  •  AC>  \  AB  •  DE  (Ax.  10). 

Now  |  AB  •  AC  =  area  A  ABC  (364), 

And  J  AB  -  DE  =  area  ABD  (?). 

Therefore  A  ABC  >  ABD  (Ax.  6). 

Q.E.D. 

This  theorem  may  be  stated  thus  :  Of  all  triangles  having 
two  given  sides,  that  triangle  whose  third  side  is  the  diameter 
of  the  circle  which  circumscribes  it  is  the  maximum. 

456.  COROLLARY.  Of  all  n-gons  having  n  —  i  sides  given, 
that  polygon  whose  nth  side  is  the  diameter  of  a  circle  which 
circumscribes  the  polygon  is  the  maximum. 


MAXIMA  AND   MINIMA  255 

PROPOSITION   XIX.    THEOREM 

457.    Of  all  isoperimetric  triangles  having  the  same  base,  the 
isosceles  triangle  is  the  maximum. 


Given :  A  ABC  and  ABD  isoper- 
imetric, having  the  same  base,  AB, 
and  A  ABC  isosceles. 

To  Prove :  A  ABC  >  A  ABD. 


Proof:  Prolong  AC  to  E,  making  CE  =  to  -4(7,  and  draw  BE. 
Using  D  as  a  center  and  BD  as  a  radius,  describe  an  arc 
cutting  EB  prolonged,  at  F.  Draw  CG  and  DH  II  to  AB,  meet- 
ing EF  at  G  and  H  respectively.  Draw  AF. 

With  C  as  a  center  and  AC,  BC  or  EC  as  a  radius,  the  Q 
described  will  pass  through  A,  B,  and  E  (Hyp.  and  Const.). 
.-.  ^ABE=rt.  Z  (240). 

That  is,  AB  is  J_  to  EF. 

Hence  CG  and  DH  are  J_  to  EF  (64). 

AC  +  CE  =  AC  +  CB  =  AD  +  DB  =  AD  +  DF 

(Hyp.  and  Const.). 

That  is,  AE=  AD  +  DF  (Ax.  1). 

But  AD  +  DF  >  AF  (Ax.  12). 

.'.  AE  >  AF  (Ax.  6). 

.'.BE>BF  (88,  IV). 

And                              i  BE  >  J  BF  (Ax.  10). 

Now                   BG  =  \  BE,  and  BH  =  I  BF  (85). 

.'.  BG  >  BH  (Ax.  6). 

Mult,  by  £  AB,    %  AB  >  BG>  %  AB  •  BH  (Ax.  10). 

But                        ±AB-BG  =  area  A  ABC  (364). 

And                      I  AB  •  BH=  area  A  ABD  (?). 

Substituting,             A  ABC  >  A  ^£D  (Ax.  6). 

Q.E.D. 


256  BOOK   V.     PLANE  GEOMETRY 

458.  COROLLARY.     Of  isoperimetric  triangles  the  equilateral 
triangle  is  the  maximum. 

[Any  side  may  be  considered  the  base.] 

PROPOSITION   XX.     THEOREM 

459.  Of  isoperimetric  polygons  having  the  same  number  of 
sides  the  maximum  is  equilateral. 

^^^^^^" ^^^^\    K/l 

Given:  Polygon  AD,  the  maximum  of 
all  polygons  having  the  same  perimeter 
and  the  same  number  of  sides. 

To  Prove :  AB  =  BC  =  CD  =  DE  =  etc. 


Proof :   Draw  AC  and  suppose  AB  not  =  to  BC. 

On  AC  as  base,  construct  A  ACM  isoperimetric  with  A  ABC 
and  isosceles  ;  that  is,  make  AM  =  CM. 

Then  A  ACM  >  A  ABC  (457). 

Add  to  each  member,  the  polygon  ACDEF. 

.*.  polygon  AMCDEF  >  polygon  AD        (Ax.  7). 
But  the  polygon  AD  is  maximum  (Hyp.). 

.-.  AB  cannot  be  unequal  to  BC  as  we  supposed  (because 
that  results  in  an  impossible  conclusion). 

Hence  AB  =  BC.     Likewise  it  is  proved  that  .8(7=  CD  =  etc. 

Q.E.D. 

460.    COROLLARY.    Of  isoperimetric  polygons  having  the  same 
number  of  sides  the  regular  polygon  is  maximum. 

Proof :    Only  one  such  polygon  is  maximum,  and  the  maxi- 
mum is  equilateral.  (459.) 
It  can  also  be  inscribed  in  a  circle  and  is  therefore  regular. 

(403.) 


MAXIMA   AND  MINIMA  257 

PROPOSITION  XXI.     THEOREM 

461.   Of  isoperimetric  regular  polygons,  the  polygon  having 
the  greatest  number  of  sides  is  maximum. 


Given:  Equilateral  A  ABC 
and  square  S,  having  the  same 
perimeter. 

To  Prove :  Square  s>  A  ABC 

AT~  ~C 

Proof:  Take  D,  any  point  in  BC,  and  draw  AD.  On  AD  as 
base,  construct  isosceles  A  ADE,  isoperimetric  with  A  ABD. 

Now  A  AED>  A  ABD  (457). 

Adding  A  ADC  to  each  member,  AEDC>  A  ABC     (Ax.  7). 
AEDCis  isoperimetric  with  A  ABC  and  S  (Hyp.  and  Const.). 

Hence  S>AEDC  (460). 

.-.  8>AABC  (Ax.  11). 

Similarly,  we  may  prove  that  an  isoperimetric  regular  pen- 
tagon is  greater  than  S ;  and  an  isoperimetric  regular  hexa- 
gon is  greater  than  this  pentagon,  etc. 

Therefore  the  regular  polygon  having  the  greatest  num- 
ber of  sides  is  maximum.  Q.E.D. 

462.  COROLLARY.  Of  all  isoperimetric  plane  figures  the  circle 
is  the  maximum. 


Ex.  1.   Of  isoperimetric  triangles,  the  maximum  is  equilateral. 

Ex.  2.    Of  all  right  triangles  that  can  be  constructed  upon  a  given 
hypotenuse,  which  is  maximum?     Why? 

Ex.  3.   Of  all  triangles  having  a  given  base  and  a  given  vertex  angle, 
the  isosceles  is  the  maximum. 

Ex.  4.   Of  all  mutually  equilateral  polygons,  that  which  can  be  in- 
scribed in  a  circle  is  the  maximum. 


258 


BOOK   V.     PLANE   GEOMETRY 


PROPOSITION  XXII.     THEOREM 

463.   Of  equal  regular  polygons  the  perimeter  of  the  polygon 
having  the  greatest  number  of  sides  is  the  minimum. 


Given :  Any  two  equal  regular  polygons,  A  and  B,  A  hav- 
ing the  greater  number  of  sides. 

To  Prove :  the  perimeter  of  A  <  the  perimeter  of  B. 

Proof :  Construct  regular  polygon  8,  similar  to  B  and  iso- 
perimetric  with  A. 

Then  A>8  (460). 

But  A  =  B  (Hyp.)- 

.-.  B>s  (Ax.  6). 

Hence    the  perimeter  of  B  >  perimeter  of  8  (376). 

But  the  perimeter  of  s  =  perimeter  of  A  (Const.). 
.*.  perimeter  of  JS>  perimeter  of  A  (Ax.  6). 

That  is,  the  perimeter  of  A<  the  perimeter  of  B.         Q.E.D. 

464.  COROLLARY.  Of  all  equal  plane  figures  the  circle  has 
the  minimum  perimeter. 

Historical  Note.  Rene*  Descartes  was  born  near  Tours,  France,  in  1596. 
He  was  a  man  of  wonderful  intellect.  He  simpli- 
fied and  generalized  the  notation  of  algebra  and  in- 
troduced the  use  of  exponents  as  now  employed. 
The  restriction  of  final  letters  of  the  alphabet  to 
represent  unknown  quantities  is  also  due  to  him. 

Descartes  was  the  first  to  adapt  algebra  to  geom- 
etry, showing  that  geometrical  figures  can  be  repre- 
sented by  algebraic  equations.  On  this  general 
truth  he  based  the  development  of  analytical  geom-  DESCARTES 

etry  which  is  known  by  his  own  name,  as  Cartesian 

geometry.     He  gave  a  large  part  of  his  life  to  original  and  creative  work 
in  mathematics,  philosophy,  physics  and  astronomy. 


ORIGINAL  EXEKCISES  259 

ORIGINAL  EXERCISES 

1.  Of  all  equal  parallelograms  having  equal  bases,  the  rectangle  has 
the  minimum  perimeter. 

2.  Of  all  lines  drawn  between  two  given  parallels  (terminating  both 
ways  in  the  parallels),  which  is  the  minimum?    Prove. 

3.  Of  all  straight  lines  that  can  be  drawn  on  the  ceiling  of  a  room 
12  feet  long  and  9  feet  wide,  what  is  the  length  of  the  maximum? 

4.  Find  the   areas  of   an    equilateral  triangle,  a  square,  a  regular 
hexagon,  and  a  circle,  the  perimeter  of  each  being  264  inches.     Which  is 
maximum  ?    What  theorem  does  this  exercise  illustrate? 

5.  Find  the  perimeters  of  an  equilateral  triangle,  a  square,  a  regular 
hexagon,  and  a  circle,  if  the  area  of  each  is  1386  square  feet.     Which 
perimeter  is  the  minimum?    What  theorem  does  this  exercise  illustrate? 

6.  Of  isoperimetric  rectangles  which  is  maximum? 

7.  Divide  a  given  line  into  two  parts  such 
that  their  product  (rectangle)  is  maximum. 

8.  Of    all  equal  triangles   having  the  same 


base,  the   isosceles  triangle    has   the  minimum        ,  ^ .^..^i. > 

perimeter. 

To   Prove:    The  perimeter  of  A  ABC  <  the 
perimeter  of  A  AB'C. 

Proof:  AD<AB'  +B'D;  etc. 
9.   Of  all  rectangles  inscribed  in. a  circle,  which  is  maximum  ?    Prove. 

10.  Of  all  rectangles  inscribed  in  a  semicircle,  which  is  maximum  ? 
Prove. 

11.  Of  all  equal  rectangles,  the  square  has  the  minimum  perimeter. 

12.  Of  all  triangles  having  a  given  base  and  a  given  vertex  angle,  the 
isosceles  triangle  has  the  maximum  area. 

13.  Of  all  triangles  having  a  given  altitude  and  a  given  vertex  angle, 
the  isosceles  triangle  is  the  minimum. 

14.  Of  all  triangles  that  can  be  inscribed  in  a  given  circle,  the  equi- 
lateral triangle  has  the  maximum  area. 

16.  The  cross  section  of  a  bee's  cell  is  a  regular  hexagon.  Would 
this  be  the  most  economical  for  the  bee  (that  is,  would  he  use  the  least 
wax)  if  one  cell  in  a  hive  were  all  he  were  to  fill  ?  Considering  also  the 
adjoining  cells,  does  the  form  of  the  regular  hexagon  require  the  least 
wax?  Explain.  Does  it  also  permit  the  storing  of  the  most  honey?  Why? 


260 


BOOK   V.     PLANE   GEOMETRY 


B 


16.  Prove,  by  the  method  employed  in  461,  that  a  regular  hexagon  is 
greater  than  an  isoperimetric  square. 

17.  Answer  the  questions  of  exercise  65  on  page  252,  without  any 
computation.     Give  reasons. 

18.  Compare  the  areas  of  the  figures  mentioned  in  Ex.  66,  page  252, 
without  performing  any  computation. 

19.  A  farmer's  house  and  barn  are  near  a  river. 
He  wishes  to  lay  from  the  house  to  the  barn,  the 
shortest  possible  path  which  shall  reach  to  the 
water's  edge.     Draw  a  plan  of  the  situation  and 
the  desired  path  and  prove  it  the  minimum. 

20.  A  farmer's  house  and  barn  are  on  opposite 
sides  of  a  straight  stream.     He  wishes  to  lay  a  road 
from  one  to  the  other,  and  erect  a  bridge  across  the 
stream  at  right  angles  to  the  banks,  by  constructing 
the  shortest  possible  track.     Draw  a  plan  of  the 
situation  and  the  desired  road,  and  prove  it  the 
minimum. 


\ 


V 


/R 


A/ 


21.   A  man  has  material  to  build  1000  yards  of 
fence.     With  this  he  desires  to  inclose  the  largest 
possible  yard  for  his  poultry.     What  will  be  the  shape  and  the  area  of 
his  yard  ? 


INDEX   OF   DEFINITIONS 


(The  numbers  refer  to  pages.) 


Abbreviations,  6. 
Acute  angle,  3. 
Acute  triangle,  5. 
Adjacent  angles,  2. 
Alternate  angles,  22. 
Alternate  exterior  angles,  22. 
Alternate  interior  angles,  22. 
Alternation,  145. 
Altitude,  of  parallelogram,  47. 

of  trapezoid,  47. 

of  triangle,  37. 
Analysis,  131. 
Angle,  2. 

acute,  3. 

bisector  of,  4,  37. 

central,  76. 

central,  of  regular  polygon,  229. 

complement  of,  3. 

degree  of,  3. 

exterior,  of  polygon,  60. 

exterior,  of  triangle,  39. 

included,  4. 

inscribed,  in  circle,  76. 

inscribed,  in  segment,  101. 

measure  of,  101. 

oblique,  3. 

obtuse,  3. 

plane,  2. 

right,  2. 

sides  of,  2. 

straight,  3. 

supplement  of,  3. 

vertex  of,  2. 
Angles,  adjacent,  2. 

alternate,  22. 

alternate  exterior,  22. 

alternate  interior,  22. 

complementary,  3. 

consecutive,  47. 

corresponding,  22. 

equal,  5. 

homologous,  of  polygons,  61. 

interior,  22. 

of  polygon,  60. 

of  quadrilateral,  47. 

of  triangle,  4. 

opposite,  of  parallelogram,  47. 


Angles  —  Continued 

opposite  interior,  39.  • 

supplementary,  3. 

vertical,  2. 
Antecedents,  143. 
Apothem,  229. 
Arc,  6,  76. 

degree  of,  101. 

intercepted,  76. 

subtended,  76. 
Arcs,  similar,  237. 
Area,  193. 
Auxiliary  lines,  19. 
Axiom,  6. 
Axioms,  6. 
Axis  of  symmetry,  65. 

Base,  of  figure,  47. 

of  triangle,  5. 
Bases,  of  parallelogram,  47. 

of  trapezoid,  47. 
Bisector  of  angle,  4,  37. 
Boundary,  1. 

Center,    figure   symmetrical   with   re- 
spect to,  65. 

of  circle,  6. 

of  regular  polygon,  229. 

of  symmetry,  65. 
Central  angle,  76. 

of  regular  polygon,  229. 
Chord,  75. 
Circle,  6,  75. 

angle  inscribed  in,  76. 

center  of,  6,  75. 

circumscribed  about  polygon,  94. 

diameter  of,  6,  75. 

inscribed  in  polygon,  94. 

radius  of,  6,  75. 

sector  of,  76. 

segment  of,  76. 

tangent  to,  75. 
Circles,  concentric,  76. 

equal,  76. 

escribed,  125. 

tangent  externally,  76. 

tangent  internally,  76. 
201 


262 


INDEX   OF   DEFINITIONS 


Circumference,  6,  75. 
Circumscribed  circle,  94. 
Circumscribed  polygon,  94. 
Commensurable  quantities,  96. 
Common  tangent,  75. 
Complement  of  angle,  3. 
Complementary  angles,  3. 
Composition,  145. 
Comppsition  and  division,  146. 
Concave  polygon,  61. 
Concentric  circles,  76. 
Conclusion,  20. 
Congruent  figures,  5. 
Congruent  polygons,  61. 
Consecutive  angles,  47. 
Consequents,  143. 
Constant,  96. 
Construction,  117. 
Continued  proportion,  143. 
Converse  of  a  theorem,  20. 
Convex  polygon,  61. 
Corollary,  8. 

Corresponding  angles,  22. 
Curved  line,  6,  75. 

Decagon,  62. 
Degree,  of  angle,  3. 

of  arc,  101. 
Demonstration,  8. 

elements  of,  20. 
Determination,  of  a  straight  line,  1. 

of  a  circle,  117. 
Diagonal,  47. 
Diameter,  6,  75. 
Direct  proportion,  166. 
Discussion,  117. 
Distance,  between  two  points,  29. 

from  a  point  to  a  line,  36. 
Division,  146. 
Dodecagon,  62. 

Equal  angles,  5. 
Equal  circles,  76. 
Equal  figures,  5. 
Equiangular  polygon,  60. 
Equiangular  triangle,  5. 
Equilateral  polygon,  60. 
Equilateral  triangle,  5. 
Escribed  circles,  125. 
Exterior  angle,  of  a  polygon,  60, 

of  a  triangle,  39. 
Extreme  and  mean  ratio,  188. 
Extremes,  143. 


Figure,  base  of,  47. 

rectilinear,  1. 

symmetrical  with  respect  to  center, 
65. 

symmetrical  with  respect  to  line,  65. 
Figures,  congruent,  5. 

equal,  5. 

isoperimetric,  254. 
Foot  of  perpendicular,  2, 
Fourth  proportional,  143. 

Geometry,  1. 
Plane,  1. 

Harmonic  division,  154. 
Heptagon,  62. 
Hexagon,  62. 
Historical  Notes : 

Archimedes,  77. 

Descartes,  258. 

Earliest  geometrical  knowledge,  8. 

Euclid,  17,  45. 

Eudoxus,  149. 

Hippocrates,  126. 

Leibnitz,  235. 

Lincoln,  217. 

Newton,  Sir  Isaac,  201. 

Plato,  131. 

Pythagoras,  85. 

Thales,  103,  160. 

Homologous  angles  in  polygons,  61. 
Homologous  parts,  5. 
Homologous  sides,  in  polygons,  61. 

in  triangles,  161. 
Hypotenuse,  5. 
Hypothesis,  19. 

Included  angles,  4. 
Incommensurable  quantities,  96. 
Indirect  method,  35. 
Inscribed  angle,  in  circle,  76. 

in  segment,  101. 
Inscribed  circle,  94. 
Inscribed  polygon,  94. 
Intercept,  to,  56,  76. 
Intercepted  arc,  76. 
Interior  angles,  22. 
Intersect,  to,  56. 
Inverse  proportion,  166. 
Inversion,  145. 
Isoperimetric  figures,  254. 
Isosceles  trapezoid,  47. 
Isosceles  triangle,  5. 


INDEX   OF   DEFINITIONS 


263 


Legs  of  isosceles  trapezoid,  47. 
Legs  of  right  triangle,  5. 
Limit  of  variable,  96,  97. 
Limits,  theorem  of,  98. 
Line,  1. 

curved,  6. 

divided  harmonically,  154. 

divided    into    extreme    and    mean 
ratio,  188. 

straight,  1. 

tangent  to  circle,  75. 
Lines,  auxiliary,  19. 

divided  proportionally,  149. 

oblique,  2. 

parallel,  3. 

perpendicular,  2. 
Locus,  114. 

Maximum,  254. 

Mean  proportional,  143. 

Means,  143. 

Measure,  of  angle,  101. 

of  quantity,  96. 

unit  of,  96. 
Median,  of  trapezoid,  47. 

of  triangle,  37. 
Method,  indirect,  35. 

of  exclusion,  35. 
Minimum,  254. 
Motion,  2. 

Mutually  equiangular  polygons,  61. 
Mutually  equilateral  polygons,  61. 

N-gon,  62. 

Oblique  angle,  3. 

Oblique  lines,  2. 

Obtuse  angle,  3. 

Obtuse  triangle,  5. 

Octagon,  62. 

Opposite  angles  of  parallelogram,  47. 

Opposite  interior  angles,  39. 

Parallel  lines,  3. 
Parallelogram,  47. 

altitude  of,  47. 

bases  of,  47. 
Parts,  homologous,  5. 
Pentagon,  62. 
Pentadecagon,  62. 
Perimeter,  94. 
Perpendicular,  2. 

foot  of,  2. 
Pi  (TT),  235. 
Plane,  1. 


Plane  angle,  2. 
Plane  geometry,  1. 
Point,  1. 

equally  distant  from  two  lines,  36. 

of  contact,  75. 
Points,  symmetrical  with  respect  to 

line  and  a  point,  65. 
Polygon,  60. 

angles  of,  60. 

apothem  of  regular,  229. 

center  of  regular,  229. 

central  angle  of  regular,  229. 

circumscribed  about  a  circle,  94. 

concave,  61. 

convex,  61. 

equiangular,  60. 

equilateral,  60. 

exterior  angle  of,  60. 

inscribed  in  a  circle,  94. 

radius  of  regular,  229. 

reentrant,  61. 

regular,  225. 

vertices  of,  60. 
Polygons,  congruent,  61. 

mutually  equiangular,  61. 

mutually  equilateral,  61. 

similar,  154. 
Pons  asinorum,  17. 
Postulate,  7. 
Problem,  117. 
Projection,  of  a  line,  170. 

of  a  point,  170. 
Proof,  8. 
Proportion,  143. 

antecedents  of,  143. 

consequents  of,  143. 

continued,  143. 

direct,  166. 

extremes  of,  143. 

inverse,  166. 

means  of,  143. 

reciprocal,  166. 
Proportional,  fourth,  143. 

mean,  143. 

third,  143. 
Proposition,  8,  117. 

Q.E.D.,  12. 
Q.E.F.,  117. 
Quadrant,  76. 
Quadrilateral,  47,  62. 

angles  of,  47. 

sides  of,  47. 

vertices  of,  47. 


264 


INDEX   OF   DEFINITIONS 


Quantities,  commensurable,  9G. 
incommensurable,  96. 

Radius,  of  circle,  6,  75. 

of  regular  polygon,  229. 
Ratio,  96,  143. 

extreme  and  mean,  188. 

series  of  equal,  143. 
Reciprocal  proportion,  166. 
Rectangle,  47. 
Rectilinear  figure,  1. 
Reductio  ad  absurdum,  35,  126. 
Reentrant  polygon,  61. 
Regular  polygon,  225. 
Rhomboid,  47. 
Rhombus,  47. 
Right  angle,  2. 
Right  triangle,  5. 

Scalene  triangle,  5. 
Secant,  75. 
Sector  of  circle,  76. 
Sectors,  similar,  237. 
Segment  of  circle,  76. 
Segments,  of  line,  149. 

similar,  237. 
Semicircle,  76. 
Series  of  equal  ratios,  143. 
Sides,  homologous,  in  polygons,  61. 

homologous,  in  triangles,  161. 

of  angle,  2. 

of  polygon,  61. 

of  quadrilateral,  47. 

of  triangle,  4. 
Similar  arcs,  237. 
Similar  polygons,  154. 
Similar  sectors,  237. 
Similar  segments,  237. 
Similar  triangles,  154. 
Solid,  1. 
Square,  47. 
Statement,  117. 
Straight  angle,  3. 
Straight  line,  1. 

divided  harmonically,  154. 

divided    into    extreme    and    mean 
ratio,  188. 

tangent  to  a  circle,  75. 
Straight  lines  divided  proportionally, 

149. 

Subtend,  to,  76. 
Subtended  arc,  76. 
Supplement  of  an  angle,  3. 


Supplementary  angles,  3. 
Surface,  1. 

area  of,  193. 

unit  of,  193. 
Symbols,  6. 
Symmetry,  65. 

axis  of,  65. 

center  of,  65. 

Tangent,  75. 

circles,  76. 
Theorem,  8. 

converse  of,  20. 

elements  of,  19. 

of  limits,  98. 
Third  proportional,  143. 
Transversal,  22. 
Trapezium,  47. 
Trapezoid,  47. 

altitude  of,  47. 

bases  of,  47. 

isosceles,  47. 

legs  of,  47. 

median  of,  47. 
Triangle,  4. 

acute,  5. 

altitude  of,  37. 

angles  of,  4. 

base  of,  5. 

equiangular,  5. 

equilateral,  5. 

exterior  angle  of,  39. 

isosceles,  5. 

median  of,  37. 

obtuse,  5. 

right,  5. 

scalene,  5. 

sides  of,  4. 

vertex  of,  5. 

vertices  of,  4. 
Triangles,  similar,  154. 

Unit,  of  measure,  96. 
of  surface,  193. 

Variable,  96. 

limit  of,  96,  97. 
Vertex,  of  an  angle,  2. 

of  a  triangle,  5. 
Vertex  angle,  5. 
Vertical  angles,  2. 
Vertices,  of  polygon,  60. 

of  quadrilateral,  47. 

of  triangle,  4. 


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